Engineering Chemistry by K. Sesha Maheswaramma, Mridula Chugh.pdf

Engineering Chemistry

K. Sesha Maheswaramma Department of Chemistry Jawaharlal Nehru Technological University, Anantapur College of Engineering Pulivendula (JNTUACEP) Andhra Pradesh

Mridula Chugh Department of Chemistry Ganga Institute of Technology and Management Jhajjar, Haryana

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Copyright © 2017Pearson India Education Services Pvt. Ltd Published by Pearson India Education Services Pvt. Ltd,CIN: U72200TN2005PTC057128, formerly known as TutorVistaGlobal Pvt. Ltd, licensee of Pearson Education in South Asia. No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 978-93-325-7118-1 eISBN 978-93-325-7879-1 Head Office: A-8 (A), 7th Floor, Knowledge Boulevard, Sector 62, Noida 201 309, Uttar Pradesh, India. Registered Office: 4th Floor, Software Block, Elnet Software City, TS 140, Block 2 & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India. Fax: 080-30461003, Phone: 080-30461060 www.pearson.co.in, Email:[emailprotected]

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Dedication To my in-laws late Smt and Sri Chavva Narayanamma and Eswara Reddy; my paternal grandparents late Smt and Sri Kalluru Akkayya and Pedda Obula Reddy; my maternal grandparents late Smt and Sri Rami Reddy Rosamma and Chinna Obula Reddy; my aunt late Smt Kalluru Nameless and my uncle late Sri Kalluru Naga Bhusana Reddy —K. Sesha Maheswaramma To my husband Gaurav Chugh for his inspiration. —Mridula Chugh

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BRIEF CONTENTS Forewordxxi Prefacexxiii About the Authors xxv

1. Water Technology

1.1

2. Polymers

2.1

3. Fuels and Combustion

3.1

4. Alternate Energy Resources

4.1

5. Electrochemistry and Batteries

5.1

6. Science of Corrosion

6.1

7. Chemistry of Engineering Materials

7.1

8. Phase Rule

8.1

9. Photochemistry

9.1

10. Surface Chemistry 10.1 11. Themodynamics

11.1

12. Metals in Biological System

12.1

13. Organometallic Compounds

13.1

14. Coordination Chemistry

14.1

15. Structure and Reactivity of Organic and Inorganic Molecules

15.1

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vi Brief Contents

16. Stereochemistry

16.1

17. Spectroscopy

17.1

18. Thermal Analysis

18.1

19. Chromatography

19.1

20. Solid State and X-Ray Diffraction

20.1

21. Green Chemistry

21.1

Lab Manual Index

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Online I.1

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CONTENTS Forewordxxi Preface xxiii About the Authors xxv 1.

Water Technology 1.1 Introduction 1.2 Sources of Water 1.3 Types of Impurities Present in Water 1.4 Hard Water and Hardness 1.5 Determination of Hardness 1.6 Dissolved Oxygen (DO) 1.7 Determination of Chlorides in Water 1.8 Determination of Acidity in Water 1.9 Alkalinity of Water 1.10 Disadvantages of Hard Water 1.11 Quality of Water for Domestic Use 1.12 Treatment of Water for Domestic Use 1.13 Break-Point Chlorination 1.14 Boilers and Boiler Troubles 1.15 Softening of Water 1.16 Desalination 1.17 Review Questions

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2. Polymers 2.1 Introduction 2.2 Degree of Polymerisation 2.3 Classification of Polymers 2.3.1 Classification Based on Source 2.3.2 Classification Based on Composition

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1.1 1.1 1.1 1.2 1.3 1.8 1.15 1.16 1.17 1.18 1.23 1.24 1.24 1.31 1.32 1.38 1.51 1.54 1.54 1.56 1.61 1.65 1.66 2.1 2.1 2.1 2.2 2.2 2.2

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2.3.3 Classification Based on Chemical Composition 2.3.4 Classification Based on Structure 2.3.5 Classification Based on Mode of Polymerisation 2.3.6 Classification Based on the Molecular Forces 2.3.7 Classification Based on Tacticity 2.4 Types of Polymerization 2.4.1 Condensation Polymerisation or Step Polymerisation 2.4.2 Addition/Vinyl/Chain Polymerisation 2.4.3 Coordination Polymerisation 2.5 Molecular Mass of a Polymer 2.6 Plastics 2.7 Important Polymers—Composition, Preparation, Properties and Engineering Uses 2.7.1 Thermoplastics 2.7.2 Thermosetting Plastics 2.8 Rubber (Elastomers) 2.8.1 Processing of Natural Rubber 2.8.2 Gutta–Percha 2.8.3 Vulcanisation of Rubber 2.8.4 Compounding of Rubber 2.8.5 Synthetic Rubbers or Artificial Rubber 2.8.6 Important Artificial Rubbers 2.9 Reinforced or Filled Plastics 2.9.1 Composition 2.9.2 Nature of Polymers Used 2.9.3 Application of Filled Plastics (Reinforced Plastics) 2.10 Biopolymers 2.10.1 Major Feed Stocks for Biopolymers 2.10.2 Preparation Methods 2.10.3 Important Biodegradable Polymers 2.10.4 Importance of Biopolymers in Sustainable Development 2.11 Conducting Polymers 2.11.1 Intrinsically Conducting Polymer (ICP) or Conjugated π-Electrons Conducting Polymer 2.11.2 Conducting Polyaniline 2.11.3 Extrinsically Conducting Polymers 2.12 Polyphosphazenes/Phosphonitrilic Polymers 2.13 Composites 2.13.1 Constituents of Composites 2.13.2 Classification of Composites 2.13.3 Advantages of Composites over Conventional Materials 2.13.4 Applications of Composites 2.14 Review Questions

2.37 2.39 2.40 2.40 2.42 2.42 2.42 2.44 2.44 2.44

2.44 2.45 2.49 2.51

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2.3 2.3 2.4 2.4 2.4 2.5 2.5 2.9 2.15 2.17 2.18 2.19 2.19 2.22 2.25 2.25 2.26 2.27 2.28 2.29 2.29 2.33 2.34 2.34 2.34 2.35 2.35 2.35 2.35 2.36 2.37

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Contents 3.

Fuels and Combustion 3.1 Introduction 3.2 Classification of Fuels 3.3 Units of Heat 3.4 Calorific Value 3.5 Determination of Calorific Value 3.5.1 Bomb Calorimeter 3.5.2 Junker’s Calorimeter 3.6 Characteristics of Good Fuel 3.7 Solid Fuels 3.7.1 Coal 3.7.2 Analysis of Coal 3.7.3 Metallurgical Coke 3.7.4 Manufacture of Metallurgical Coke 3.8 Liquid Fuels 3.8.1 Petroleum Refining 3.8.2 Important Petroleum Products and their Uses 3.9 Synthetic Petrol 3.9.1 Cracking 3.9.2 Fischer–Trapsch Method 3.9.3 Bergius Method 3.10 Power Alcohol 3.10.1 Manufacture of Power Alcohol 3.11 Knocking 3.12 Diesel Engine, Cetane and Octane Number 3.13 Gaseous Fuels 3.13.1 Natural Gas 3.13.2 Producer Gas (or) Suction Gas 3.13.3 Water Gas (or) Blue Gas 3.13.4 Coal Gas 3.13.5 Biogas 3.14 Flue Gas Analysis by Orsats Apparatus 3.15 Review Questions

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4. Alternate Energy Resources 4.1 Introduction 4.1.1 Conventional or Traditional Energy Resources 4.1.2 Nonconventional Energy Resources or Renewable Energy Sources 4.1.3 Alternative Energy 4.2 Non-Conventional Energy Sources and Storage Devices 4.2.1 Solar Energy 4.2.2 Wind Energy 4.2.3 Geothermal Energy

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ix 3.1 3.1 3.1 3.2 3.2 3.3 3.3 3.4 3.7 3.8 3.8 3.8 3.16 3.17 3.18 3.18 3.20 3.21 3.21 3.23 3.23 3.24 3.25 3.26 3.26 3.28 3.28 3.29 3.29 3.30 3.30 3.31 3.32 3.32 3.33 3.37 3.38 4.1 4.1 4.1 4.1 4.2 4.2 4.2 4.9 4.10

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x Contents

4.2.4 Water Power 4.2.5 Biomass 4.2.6 Nuclear Energy 4.2.7 Nuclear Reactions 4.3 Review Questions

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Electrochemistry and Batteries 5.1 Introduction 5.2 Electrolysis 5.2.1 Laws of Electrolysis 5.3 Electrolytic Conduction 5.3.1 Factors Affecting Electrolytic Conduction 5.3.2 Electrical Resistance and Conductance 5.3.3 Specific, Equivalent and Molar Conductivities 5.3.4 Equivalent Conductivity 5.3.5 Molar Conductivity 5.3.6 Measurement of Electrolytic Conductance 5.3.7 Variation of Conductivity with Concentration 5.3.8 Conductance Behaviour of Strong Electrolyte 5.3.9 Conductance Behaviour of Weak Electrolyte 5.4 Kohlrausch’s Law of Independent Migration of Ions 5.5 Conductometric Titrations 5.6 Electrochemical Cells 5.7 Types of Electrodes 5.8 Reference Electrode 5.9 Ion Selective Electrodes (ISE) 5.9.1 Electrochemical Circuit and Working of ISE 5.9.2 Types of Ion – Selective Membranes 5.9.3 Applications of Ion Selective Electrodes 5.10 Glass Electrode 5.10.1 Construction of Glass Electrode 5.11 Concentration Cell 5.12 Potentiometric Titrations 5.13 Electrochemical Sensors 5.13.1 Potentiometric Sensor 5.13.2 Analysis of Glucose in Blood 5.13.3 Analysis of Urea 5.14 Voltammetry 5.14.1 Linear Sweep Voltammetry (LSV) 5.14.2 Ferric Fe3+/Fe2+ System 5.14.3 Cyclic Voltammetry 5.14.4 Applications of Voltammetry

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4.10 4.12 4.13 4.15 4.23 4.23 4.24 4.26 4.26 5.1 5.1 5.1 5.2 5.4 5.4 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.10 5.11 5.15 5.21 5.28 5.31 5.34 5.34 5.35 5.37 5.38 5.38 5.40 5.42 5.44 5.44 5.45 5.45 5.45 5.46 5.47 5.49 5.50

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Contents

5.15 Batteries 5.15.1 Advantages of Batteries 5.15.2 Disadvantages of Batteries 5.16 Review Questions

5.51 5.51 5.51 5.60

5.60 5.62 5.66 5.71 5.78 5.79

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6. Science of Corrosion 6.1 Introduction 6.1.1 Causes of Corrosion 6.1.2 Types of Corrosion 6.2 Galvanic Series 6.2.1 Factors Affecting Corrosion 6.3 Protection from Corrosion (Preventive Measures for Corrosion Control) 6.4 Review Questions

6.1 6.1 6.1 6.2 6.16 6.17 6.19 6.29

6.29 6.31 6.35 6.38

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7. Chemistry of Engineering Materials 7.1 Semiconducting and Super Conducting Materials 7.1.1 Semiconductor 7.1.2 Applications of Semiconductors 7.1.3 Superconductors 7.2 Magnetic Materials 7.2.1 General Properties of Magnetic Materials 7.2.2 Classification of Magnetic Materials 7.2.3 Applications of Magnetic Materials 7.3 Cement 7.3.1 Classification of Cement 7.3.2 Raw Materials used in the Manufacture of Portland Cement 7.3.3 Manufacture of Portland Cement 7.3.4 Chemical Composition of Portland Cement and its Importance 7.3.5 Setting and Hardening of Cement 7.3.6 ISI Specifications of Cement 7.3.7 Analysis of Cement 7.3.8 Plaster of Paris/Gypsum Plaster 7.4 Refractories 7.4.1 Characteristics of Good Refractory Materials 7.4.2 Failures of Refractory Materials 7.4.3 Classification of Refractories

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7.1 7.1 7.1 7.3 7.3 7.4 7.4 7.4 7.6 7.6 7.7 7.7 7.8 7.12 7.13 7.15 7.16 7.16 7.17 7.17 7.18 7.18

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7.4.4 Properties of Refractories 7.4.5 Manufacture of High-Alumina Bricks, Magnesite Bricks and Zirconia Bricks 7.5 Lubricants 7.5.1 Important Functions of Lubricants 7.5.2 Mechanism of Lubrication 7.5.3 Classification of Lubricants 7.5.4 Properties of Lubricants 7.5.5 Redwood Viscometer 7.5.6 Engler’s Viscometer 7.5.7 Saybolt Viscometer 7.5.8 U-Tube Viscometer 7.5.9 Conversion of Redwood, Engler and Saybolt Viscosities into Absolute Units 7.6 Explosives and Propellants 7.6.1 Some Important Terms about Explosives 7.6.2 Classification of Explosives 7.6.3 Precautions during Storage of Explosives 7.6.4 Blasting Fuses 7.6.5 Important Explosives and their Preparation 7.6.6 Rocket Propellants 7.6.7 Characteristics of a Good Propellant 7.6.8 Classifications of Propellants 7.7 Nanomaterials 7.7.1 Synthesis of Nanomaterials 7.7.2 Characterisation 7.7.3 Importance 7.7.4 Broad Classification of Nanomaterials 7.7.5 Fullerenes 7.7.6 Types of Fullerenes 7.7.7 Properties of Nanomaterials 7.7.8 Applications of Nanomaterials 7.8 Liquid Crystals 7.8.1 Characteristics of Liquid Crystal Phase 7.8.2 Classification of Liquid Crystals 7.8.3 Thermotropic Liquid Crystals 7.8.4 Lyotropic Liquid Crystals 7.8.5 Chemical Properties of Liquid Crystals 7.8.6 Applications of Liquid Crystals 7.9 Abrasives 7.9.1 Hardness of Abrasive 7.9.2 Natural Abrasives 7.9.3 Artificial Abrasives 7.10 Review Questions

7.35 7.36 7.36 7.37 7.39 7.40 7.40 7.41 7.42 7.43 7.44 7.44 7.45 7.45 7.45 7.46 7.46 7.49 7.51 7.51 7.52 7.52 7.52 7.57 7.57 7.58 7.59 7.59 7.60 7.60 7.61

7.61 7.64 7.76 7.82

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7.18 7.21 7.22 7.22 7.23 7.24 7.27 7.31 7.33 7.33 7.34

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Contents

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8. Phase Rule 8.1 8.1 Introduction 8.1 8.2 Explanation of the Terms Involved in Phase Equilibria 8.1 8.2.1 Phase (P) 8.1 8.2.2 Components (C) 8.2 8.2.3 Degree of Freedom (F) 8.4 8.2.4 True and Metastable Equilibrium 8.5 8.2.5 Eutectic Mixture and Eutectic Point 8.5 8.2.6 Triple Point 8.6 8.3 Phase Rule 8.6 8.3.1 Assumptions for the Validation of Phase Rule 8.6 8.3.2 Thermodynamic Derivation of the Phase Rule 8.6 8.3.3 Utility of Phase Rule | Application of Phase Rule 8.8 8.3.4 Limitations of Phase Rule 8.8 8.4 Phase Diagrams 8.9 8.5 One Component System 8.9 8.6 Two Component System 8.14 8.6.1 Eutectic System 8.14 8.6.2 Lead (Pb) – Silver (Ag) System 8.18 8.7 Heat Treatment of Steel 8.21 8.8 Review Questions 8.22

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8.22 8.24 8.28 8.32 8.34

9. Photochemistry 9.1 Introduction 9.2 Light Source in Photochemistry 9.3 Laws of Photochemistry 9.3.1 Grotthuss–Draper Law or The First Law of Photochemistry 9.3.2 Stark-Einstein Law or Photochemical Equivalence Law 9.3.3 Beer-Lambert Law 9.4 Photophysical and Chemical Processes 9.4.1 Photophysical Process 9.4.2 Photochemical Process 9.5 Quantum Yield and Quantum Efficiency 9.6 Photosensitisation 9.7 Photodynamic Therapy 9.8 Important Photochemical Reactions 9.9 Review Questions

9.1 9.1 9.1 9.2 9.2 9.2 9.3 9.3 9.3 9.4 9.6 9.6 9.6 9.7 9.9

9.9 9.10 9.10 9.11

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xiv Contents 10. Surface Chemistry 10.1 Introduction 10.2 Adsorption 10.2.1 Mechanism of Adsorption 10.2.2 Adsorption is Exothermic 10.2.3 Difference between Adsorption and Absorption 10.2.4 Examples of Adsorption, Absorption, and Sorption 10.2.5 Positive and Negative Adsorptions 10.2.6 Classification of Adsorption 10.2.7 Factors Affecting the Adsorption of Gases by Solids 10.2.8 Adsorption Isotherms 10.2.9 Applications of Adsorption 10.3 Colloidal State 10.3.1 Types of Solution 10.3.2 Classification of Colloids 10.3.3 Properties of Colloidal Solutions 10.3.4 Applications of Colloids 10.4 Review Questions

10.1 10.1 10.1 10.2 10.2 10.2 10.3 10.3 10.3 10.4 10.5 10.16 10.17 10.17 10.18 10.21 10.28 10.31

10.31 10.33 10.37 10.39

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11. Thermodynamics 11.1 Introduction 11.1.1 Thermodynamic Terms and Basic Concepts 11.2 Types of Thermodynamic Systems 11.2.1 Isolated System 11.2.2 Closed System 11.2.3 Open System 11.3 Intensive and Extensive Properties 11.3.1 Intensive Property 11.3.2 Extensive Property 11.3.3 State Variables 11.4 Reversible and Irreversible Process 11.4.1 Reversible Process 11.4.2 Irreversible Process 11.4.3 Thermodynamic Processes 11.4.4 Isothermal Process or Isothermal Change 11.4.5 Indicator Diagram 11.4.6 Work Done by a System in an Adiabatic Process 11.4.7 First Law of Thermodynamics and its Application 11.4.8 Second Law of Thermodynamics 11.4.9 Carnot’s Engine, Efficiency 11.4.10 Working of Carnot’s Engine 11.4.11 Absolute Zero 11.4.12 Numerical Problems Based on Carnot’s Cycle

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11.1 11.1 11.1 11.2 11.2 11.2 11.2 11.3 11.3 11.3 11.3 11.3 11.3 11.4 11.4 11.5 11.6 11.8 11.10 11.13 11.14 11.16 11.18 11.19

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Contents

11.4.13 Solved Numerical Problems Based on Isothermal and Adiabatic Process 11.21 11.5 Thermodynamic Potentials and Maxwell Equations 11.33 11.5.1 Thermodynamic Potential 11.33 11.5.2 Internal Energy (U)11.33 11.5.3 Total Heat Function (H)11.34 11.5.4 Helmholtz Function (F)11.35 11.5.5 Gibb’s Free Energy or Gibb’s Function (G)11.36 11.5.6 Maxwell’s Equations 11.37 11.5.7 Clausius–Clapeyron Equation 11.38 11.5.8 Derivation of the Stefan–Boltzmann Law using Maxwell’s Equations 11.40 11.5.9 Joule–Thomson Effect or Joule–Kelvin Effect 11.44 11.6 Review Questions 11.48

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11.48 11.49 11.50 11.52

12. Metals in Biological System 12.1 12.1 Introduction 12.1 12.2 Essential Elements 12.1 12.2.1 Bulk Elements 12.2 12.2.2 Macrominerals 12.2 12.2.3 Micro Elements (Trace Elements) 12.2 12.3 Non-Essential Elements 12.4 12.4 Important Metals in Biological Systems 12.4 12.4.1 Haemoglobin 12.4 12.4.2 Myoglobin 12.6 12.4.3 Vitamin B1212.7 12.4.4 Chlorophyll 12.8 12.5 Metals and their Toxicity 12.9 12.5.1 Toxicity of Arsenic 12.9 12.5.2 Toxicity of Lead 12.10 12.5.3 Toxicity of Mercury 12.11 12.6 Review Questions 12.11

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13. Organometallic Compounds 13.1 Introduction 13.1.1 Organometallic Chemistry Timeline 13.2 Organolithium Compounds 13.2.1 Preparation of Organolithium Compounds 13.2.2 Properties of Organolithium Compounds

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12.11 12.12 12.13 12.14 13.1 13.1 13.1 13.2 13.2 13.3

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13.3 Organomagnesium Compounds 13.3.1 Preparation of Organomagnesium Compounds 13.3.2 Properties of Organomagnesium Compounds 13.4 Metal Carbonyls 13.4.1 Ligand 13.4.2 Effective Atomic Number 13.4.3 Preparation of Carbonyls 13.4.4 Properties of Carbonyls 13.4.5 Structure of Carbonyls 13.5 Review Questions

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13.5 13.5 13.7 13.10 13.11 13.11 13.11 13.12 13.12 13.15 13.15 13.15 13.16 13.16

14. Coordination Chemistry 14.1 Introduction 14.2 Basic Requirements to Formation of Coordination Compound 14.3 Nomenclature of Metal Complexes 14.3.1 Cationic Complex 14.3.2 Anionic Complex 14.3.3 Nonionic Complexes 14.3.4 Polynuclear Complex 14.3.5 Complex with Metal-Metal Bond 14.4 Theories of Coordination Chemistry 14.4.1 Werner’s Theory 14.4.2 Sidgwick’s Electronic Concept Theory 14.4.3 Valance Bond Theory 14.4.4 Crystal Field Theory 14.4.5 Common Single Atomic Ligands and their Field Strength 14.4.6 Molecular Orbital Theory of Coordination Complexes 14.5 Factors Affecting the Stability of Coordination Compounds 14.6 Determination of Complex Ion Formation 14.7 Stability of Coordination Compounds 14.8 Applications of Coordination Compounds 14.9 Review Questions

14.1 14.1 14.1 14.3 14.4 14.4 14.5 14.6 14.6 14.7 14.7 14.8 14.9 14.11 14.14 14.16 14.17 14.19 14.20 14.22 14.23

14.23 14.25 14.27 14.28

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15. Structure and Reactivity of Organic and Inorganic Molecules 15.1 15.1 Introduction 15.1 15.2 Hybridisation 15.1 15.2.1 Salient Features of Hybridisation 15.1 15.2.2 Important Conditions for Hybridisation 15.2 15.2.3 Types of Hybridisation 15.2

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15.3 Bond Polarisation 15.5 15.3.1 Electron Displacement in Covalent Bonds 15.5 15.4 Reaction Intermediates 15.9 15.4.1 Free Radicals 15.9 15.4.2 Carbocations or Carbonium Ions 15.10 15.4.3 Carbanions 15.12 15.4.4 Carbenes 15.13 15.4.5 Nitrenes or Imidogens 15.13 15.4.6 Benzynes 15.14 15.5 Molecular Orbital Theory 15.19 15.5.1 Important Points on Molecular Orbital Diagrams 15.21 15.5.2 Fundamental Steps for Constructing Molecular Orbitals 15.21 15.5.3 Five Basic Rules of Molecular Orbital Theory 15.22 15.5.4 Linear Combination of Atomic Orbitals and Type of Atomic Orbitals 15.22 15.5.5 Molecular Orbital Energy Level Diagrams of Homo Atomic Molecules15.23 15.5.6 Molecular Energy Level Diagrams of Hetero Atomic Molecules 15.29 15.6 Review Questions 15.33

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16. Stereochemistry 16.1 Introduction 16.2 Isomerism 16.2.1 Structural Isomerism 16.2.2 Space or Stereoisomerism 16.3 Classification of Structural Isomerism 16.3.1 Chain or Nuclear Isomerism 16.3.2 Position Isomerism 16.3.3 Ring or Chain Isomerism 16.3.4 Functional Group Isomerism 16.3.5 Metamerism 16.3.6 Tautomerism 16.4 Classification of Stereoisomerism 16.4.1 Geometrical Isomerism 16.4.2 Optical Isomerism 16.4.3 Conformational Isomers 16.4.4 R–S Nomenclature or CIP Nomenclature 16.4.5 E–Z Nomenclature 16.5 Molecular Representation 16.5.1 Wedge and Dash Projections 16.5.2 Fisher Projections 16.5.3 Sawhorse Representation 16.5.4 Newman Representation 16.6 Molecular Isomerism 16.7 Review Questions

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15.33 15.34 15.36 15.38 16.1 16.1 16.1 16.1 16.1 16.1 16.2 16.2 16.2 16.2 16.2 16.3 16.3 16.3 16.4 16.10 16.10 16.12 16.12 16.12 16.13 16.13 16.14 16.14 16.15

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17. Spectroscopy 17.1 Introduction 17.2 Ultra Violet and Visible Spectroscopy 17.2.1 Principle 17.2.2 Instrumentation 17.2.3 Instrumental Design 17.2.4 Electronic Transitions 17.2.5 Chromophores 17.2.6 Auxochrome 17.2.7 Woodward–Fieser Rules 17.2.8 Factors Affecting the Position of the λ Maximum and Intensity of Radiation 17.2.9 Franck-Condon Principle 17.2.10 Solved Problems Based on UV-Vis Spectroscopy 17.2.11 Applications of UV-Visible Spectroscopy 17.3 IR-Spectroscopy 17.3.1 Basic Principle 17.3.2 Instrumentation 17.3.3 Molecular Vibrations 17.3.4 Factors Affecting Vibrational Frequency 17.3.5 Degrees of Freedom 17.3.6 Solved Problems Based on IR Spectra 17.3.7 Applications of IR Spectroscopy 17.4 Nuclear Magnetic Resonance Spectroscopy 17.4.1 Principle 17.4.2 Instrumentation 17.4.3 Chemical Shift 17.4.4 Spin-Spin Splitting, Spin-Spin Interaction, Spin–Spin Coupling or Fine Spectrum 17.4.5 Magnetic Resonance Imaging 17.4.6 High Resolution Proton Magnetic Resonance Spectroscopy 17.4.7 NMR Applications 17.4.8 Solved Problems Based on Proton NMR 17.5 Review Questions

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18. Thermal Analysis 18.1 Introduction 18.2 Thermogravimetric Analysis

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16.15 16.16 16.18 16.19 17.1 17.1 17.2 17.2 17.2 17.3 17.5 17.6 17.6 17.7 17.10 17.11 17.12 17.13 17.14 17.15 17.16 17.17 17.22 17.22 17.22 17.23 17.24 17.24 17.25 17.26 17.28 17.29 17.30 17.32 17.33 17.36 17.36 17.37 17.39 17.41 18.1 18.1 18.1

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18.2.1 Principle of TGA 18.2.2 Applications of TGA 18.3 Differential Thermal Analysis 18.3.1 Principle of DTA 18.4 Review Questions

18.2 18.3 18.4 18.4 18.5

18.5 18.5 18.6 18.6

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19. Chromatography 19.1 19.1 Introduction 19.1 19.1.1 Chromatography Timeline 19.1 19.2 Classification of Chromatography 19.2 19.2.1 Classification Based on Mobile Phase 19.2 19.2.2 Classification Based on Attractive Forces 19.2 19.2.3 Classification Based on Partition of Relative Solubility of Analyte in Mobile and Stationary Phase 19.2 19.2.4 Chromatographic Techniques on the Type of Support Material Used in the System 19.2 19.3 Types of Chromatography 19.3 19.3.1 Gas-Liquid-Chromatography 19.3 19.4 Chromatography Theory 19.10 19.4.1 Distribution Coefficient or Partition Coefficient (K)19.10 19.4.2 Retention Time (tR)19.10 19.4.3 Retention Volume (VR)19.11 19.4.4 Plate Theory 19.12 19.5 High Performance Liquid Chromatography 19.14 19.5.1 Instrumentation 19.15 19.5.2 Theory of High Performance Liquid Chromatography 19.19 19.6 Review Questions 19.26

Fill in the Blanks Multiple-choice Questions Short Answer Questions Descriptive Questions

20. Solid State and X-Ray Diffraction 20.1 Introduction 20.1.1 Crystal Structure 20.2 Crystal Systems 20.2.1 Laws of Crystallography 20.3 Crystal Defects 20.3.1 Stoichiometric Defect 20.3.2 Non-stoichiometric Defect 20.4 X-Ray Diffraction 20.4.1 Introduction 20.4.2 Principle

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19.26 19.27 19.27 19.29 20.1 20.1 20.1 20.1 20.2 20.3 20.3 20.4 20.6 20.6 20.6

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xx Contents

20.4.3 X-ray Diffraction of Crystals and Bragg’s Equation 20.4.4 Determination of Crystal Structure with Bragg’s Equation 20.4.5 X-ray Diffraction Methods 20.4.6 Instrumentation of X-ray 20.5 Application of X-ray Diffraction 20.6 Review Questions

Fill in the Blanks Multiple-choice Questions Short Answer Questions Descriptive Questions

20.6 20.8 20.8 20.9 20.10 20.10 20.10 20.11 20.11 20.12

21. Green Chemistry 21.1 Introduction 21.2 Twelve Principles of Green Chemistry 21.3 Importance of Green Synthesis 21.3.1 Methods for Green Synthesis 21.3.2 Applications of Green Synthesis 21.4 Greenhouse Concepts 21.4.1 Types of Greenhouse 21.5 Greenhouse Gases and Greenhouse Effect 21.5.1 Natural Greenhouse Effect 21.5.2 Enhanced Greenhouse Effect 21.5.3 Greenhouse Gas Effect 21.5.4 Requirements for Greenhouse 21.6 Carbon Sequestration 21.6.1 Importance of Carbon Sequestration 21.7 Why Carbon Dioxide is a Major Problem 21.8 Review Questions

21.1 21.1 21.1 21.2 21.3 21.4 21.5 21.5 21.5 21.6 21.6 21.7 21.7 21.8 21.8 21.9 21.9

21.9 21.9 21.10 21.10

Fill in the Blanks Multiple-choice Questions Short answer Questions Descriptive Questions

Lab Manual Index

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Online I.1

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FOREWORD I feel proud to write these few lines as my student Smt K Sesha Maheswaramma, working as assistant professor, Department of Chemistry, JNTUACEP, Andhra Pradesh, has brought out this book successfully. With a personal note, I wish to place on records that from a remote village Gollala Guduru of Kadapa District, where the expectation of girl child’s higher education is minimal, she not only became the assistant professor of an engineering college but also started publishing books! Although the subject Chemistry is unwelcomed by the students of this region, she has mastered not only the subject but also the art of writing. In the recent past, with too much specialization, it was very difficult for anyone to have a comprehensive understanding of any subject with broader vision due to limited resources to teaching and research. To overcome this, recently, ‘Interdisciplinary Approach’ is stressed very much. In this context, Dr K. Sesha Maheswaramma has brought out Engineering Chemistry by integrating Chemistry with water technology, alternate energy resources, engineering materials, thermodynamics, analytical methods, green chemistry, etc. This book is most useful to all undergraduate students (both B.Tech. and B.Sc.) and postgraduate students of M.Sc. Chemistry as well as M.Sc. Physics. It is true that scientific knowledge is that of empirical—dealing with facts. Reporting of facts done through the medium of language is marvellous. I sincerely appreciate her hard work and efforts in bringing out this book.

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Rev. Fr. Dr. T. Amala Arockia Raj, SJ Principal, Loyola Degree College (YSRR) Pulivendula, Kadapa (Dt.), Andhra Pradesh

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PREFACE Chemistry, as a basic science, plays a vital role in making a foundation for engineering students and ensuring security by shaping a fruitful superstructure for their entire career. A comprehensive knowledge of fundamental chemistry is essential for promising and sustainable future. Chemistry plays a leading role in shaping the future of research and development with the integration of multidisciplinary fields such as life sciences, material science, medicine, engineering and technology. Engineering Chemistry text book is exclusively designed for the needs of undergraduate students of all disciplines of science, engineering and technology. This book is designed based on the syllabi of various universities and autonomous colleges of India. Hence, this book highly recommended for both graduate and under graduate students of science, engineering and technology. This book introduces the fundamental concepts in a simple and illustrative manner. This book can also be a useful as selflearning guide for students as well as teachers to teach in a highly practical way. This book contains 21 theory chapters and 14 lab experiments as an online supplement as per the latest engineering chemistry syllabi offered in all Indian Universities. Each chapter starts with brief introduction and provides indepth information of the relevant topics covered in the syllabus. The textbook covers all the topics and includes latest information with diagrams, tables and solved numerical problems, and the review zone contains fill in the blanks, multiple-choice questions, short answer questions and descriptive questions for examinations. As the book discusses the fundamentals with a practical approach it is suggested for several nationwide competitive exams such as NET, SET, GATE Groups and Civil Services. We heartily welcome valuable comments and suggestions from our readers for the improvement of the future editions of this book, which may be addressed to [emailprotected] and [emailprotected].

ACKNOWLEDGEMENTS From the bottom of my heart, I give all glory and praises to the Almighty for his everlasting mercy and abundant compassion to accomplish this work. Mere words could never express my gratitude to my teachers who provided me all support, guidance and precious life lessons. They activated hunger for knowledge and wisdom, and inspired me to plan a better future by adding my brick of effort to the world of Chemistry. I am beholden to the authorities of JNTUA, Anantapur, for their encouragement to sharpen my skills and contribute same to the development of student community. This effort of ours would not have been a success but for the contributions made by a lot of people.

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xxiv Preface I express my heartfelt thanks to the members of editorial and production team of Pearson Education, especially Mr Sojan Jose, Ms R. Dheepika and Mr M. Balakrishnan, for their unrelenting efforts in editing the manuscript and their immense contribution has shaped this stupendous work. I wish to express my sincere thanks to my family members. I owe much to my parents Sri Kalluru Naga Malla Reddy and Smt. Gangulamma, my brothers Sri Mallikarjuna Reddy and Sri Nagarjuna Reddy for their cherished, unstinted love and encouragement in all my good activities. I wish to express my indebted thanks to my husband Dr C. Mallikarjuna Reddy for his unbounding affection and source of constant inspiration throughout my work. I am grateful to my sister-in-law Smt. T. Chandra Leelavathi and my brother Sri T. Raja Sekhara Reddy, for their perpetual dedication to carry out my family commitments with all their moral and physical support while I struggled with my work. My hearty cheers to my beloved children C. Hitesh Reddy and C. Hasya Reddy, whose love and smile encouraged me in completing this work. I am thankful to my colleagues, friends and students who helped me a lot to write this book.

K. Sesha Maheswaramma

I express my thanks to all of my family members. First and foremost, I would like to thank my h usband Gaurav Chugh for standing by me throughout my career and supporting me in writing this book. He is a continuous source of inspiration. Iwould like to thank my parents and my siblings for allowing me to follow my ambitions throughout my childhood. My family, including my in-laws, has always supported me throughout my career and authoring this book and I really appreciate it. I would also like to thank Dr Y. Madhveelatha, Principal, MRECW for her encouragement and Dr Pawan Sharma for suggestions and guidance. I also owe my thanks to my well-wishers Dr Nidhi Dureja, Dr Gaurav Dhingra, Dr Saritha and Ms Tanuja Safala and Ms Harshitha. Mridula Chugh

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ABOUT THE AUTHORS K. Sesha Maheswaramma is presently assistant professor in the Department of Chemistry, JNTUA College of Engineering, Pulivendula, YSR Kadapa (Dt.), Andhra Pradesh. She has done her M.Sc. in Mineral Chemistry, M.Phil. in Water treatment and Ph.D. in Solid State Chemistry from Sri Venkateswara University, Tirupathi, Andhra Pradesh. She won first rank in M.Sc. and was awarded Justice P. C. Reddy meritorious gold medal in the year 2001 at the university level. She has rich privilege of gaining fifteen years of teaching experience for both graduate and post-graduate students of B.Sc., M.Sc. and B.Tech. She has zeal for research and established herself as a good guide to encourage students for challenging areas of research. Her research interests spread over analytical chemistry, bio-inorganic medicinal chemistry, solid state spectroscopy and environmental challenges. She has more than ten research publications in internationally reputed journals and presented more than twenty research papers in national and international conferences, which are widely acclaimed as the most relevant for addressing the existing societal problems and discussing practical remedies. Maheswaramma is a coordinator, organizer, speaker, invitee and facilitator for several regional and national level seminars and training programmes. She is expertise in academic administration, framing academic guidelines, regulations, syllabi, coordinating various boards like studies, academic, counseling, career, result analysis and active in chairing debates and discussions. She visited GFZ Research Centre, Potsdam, Berlin, Germany, and Chinese National Academy of Science, Beijing, China, and presented research papers. She is a life member of Indian Society of Technical Education (ISTE) and awarded the IASc–INSA–NASI research fellowship for the years 2013 and 2014. Mridula Chugh is an assistant professor in Ganga Institute of Technology and Management, Jhajjar, Haryana. Previously, she worked with Malla Reddy Engineering College for Women, JNTUH for three years. She has pursued M.Sc. (Organic Chemistry), M.Phil., and CSIR UGC-NET. She did her master’s degree in Kurukshetra University, Kurukshetra. She has a teaching experience of ten years. She has attended various national and international seminars and conferences and presented papers.

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WATER TECHNOLOGY Water is the driving force in nature, we never know the worth of water till the river is dry.

1.1

INTRODUCTION

Water is a natural wonder and is the most common, important, useful thing for surviving of all the living beings. Without food, living beings can survive for some days but without water nobody can survive. Seventy percent of our body contains water, which regulates life processes such as digestion of food, transportation of nutrients, and excretion of body wastes. It regulates the body temperature by the process of sweating and evaporation. Water acts as a universal solvent; due to this reason, water is widely used in laboratories, irrigation, steam generator, industrial purpose, fi re fighting, etc. Besides it is used for bathing, drinking, sanitary purposes, etc. From an engineer’s point of view too, very important, without water nothing will happen. It is required in boilers for production of steam, which acts as a source of energy and a coolant in many power and chemical plants and many other industries.

1.2

SOURCES OF WATER

Water present on earth passes through a remarkable cycle of changes (as shown in Figure 1.1): (i) Rain water is the purest form of natural water because it is obtained by evaporation from the surface water. But rain water during its downward path to the earth dissolved considerable amount of gases (e.g. CO2, NO, NO2, SO2, SO3 etc) and suspended solid particles, which are present in the atmosphere. So it becomes polluted. (ii) The water that comes to the surface through rain is in the form of river water and lake water. River water contains dissolved inorganic salts such as chlorides as well as dissolved impurities from the soil. In lake water, the main impurities are organic matter. (iii) A larger amount of rain water is percolated in the soil through permeable rocks, loose sand, gravel, etc. During its passage downward into the ground, the suspended matter is left behind, and organic matter is oxidized by bacteria. This water is extremely clear as a result of natural filtration through the sand bed.

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1.2

Engineering Chemistry Sources

Sea water (approx 97%)

Fresh water (approx 3%)

Surface water

Rain water

Underground water

Flowing water Still water Streams Ponds Rivers

Lakes Reservoirs

Springs Wells Tubewells

Figure 1.1 Flow diagram of sources of water (iv) Sea water is the most impure form of natural water. River water joins the sea and thus gives its impurities to the sea. In addition, due to evaporation of water, sea is having about 3.5% of dissolved salts, and the maximum amount (2.6%) is due to sodium chloride. Hence sea water is salty in taste. This water is not directly useful to man because it is not palatable as it contains 2000 times more dissolved salt than fresh water.

1.3

TYPES OF IMPURITIES PRESENT IN WATER

Water may contain various impurities due to (i) (ii) (iii) (iv)

The ground or soil with which it comes in contact (e.g. garbage, soil particles, etc.) Its contact with sewage or industrial wastes The decomposition of dead plants and animals The growth of bacteria, algae, viruses, etc.

The common impurities present in natural water can be classified into four groups that are as follows and shown in Figure 1.2: (i) Dissolved impurities (a) Dissolved gases – NO2, CO2, SO2, etc., which are soluble in water and make it impure. (b) Dissolved inorganic salts or ions (1) Cations: Ca2+, Mg2+, Na+, K+, Fe2+, Al3++, Zn2+, etc. (2) Anions: CO32− , SO2− , HCO3− , Cl−, etc. 4 (ii) Suspended impurities (a) Inorganic – sand, clay, lime, etc. (b) Organic – Plant and animal materials like discarded vegetables, dry leaves, dead materials,etc.

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Water Technology

1.3

Types of Impurities

Dissolved

Suspended

Inorganic

Organic

Gases

Inorganic

Colloidal

Microorganism

Figure 1.2 Types of impurities (iii) Colloidal impurities Finely divided silica, clay, organic products, colouring matter, etc. (iv) Microorganism Various pathogenic microorganisms such as bacteria, fungi, virus, etc. The various types of impurities present in the water impact certain properties in water. (a) Presence of different chemicals impart colour, odour and taste to the water. (b) Presence of dissolved salt makes the water hard. (c) Excess quantities of metals and dissolved gases make the water corrosive in nature. (d) Presence of pathogenic bacteria in water makes it unfit for drinking or domestic purposes. (e) Suspended matter create turbidity to the water.

1.4

HARD WATER AND HARDNESS

Depending on salts presents in water and reaction with soap, water is categorized into hard water and soft water. Hardness is the characteristic of water by which water does not produce lather with soap. It is due to presence of chlorides, sulphates and bicarbonates salts of magnesium, calcium and other heavy metals [CaCl2, CaSO4, MgCl2, MgSO4, Ca(HCO3)2, Mg(HCO3)2, etc]. When hard water is treated with soap, it does not produce lather, rather it forms a white scum. Soap is the sodium or potassium salt of higher fatty acids like stearic acid [C17H35 COONa – sodium stearate]. 2C17 H 35 COONa + MgCl 2  → (C1177 H 3355 COO)2 Mg + 2NaCl Sodium stearate

Salt

Insoluble magnesium stearate/white scum

2C17 H 35 COOK + MgSO4    → (C17 COO O)2 Mg + K 2SO 4 17 H 3355 CO Pottasium stearate

Saallt

White scum

2C17 H 35 COONa + CaCl 2  → (C1177 H 3355 COO)2 Ca + 2 NaCl Sodium stearate

Salt

White scum

2C17 H 35 COONaa + CaS CaSO C aSO O 4  → (C1177 H 35 COO OO)2 Ca + Na 2SO 4 35 C Soap

Salt

White scum

Water which can produces lather with soap easily is called soft water.

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1.4

Engineering Chemistry

(i) Types of hardness Depending on salts present in the water, hardness is of two types, i.e., temporary hardness and permanent hardness. (a) Temporary or Carbonate Hardness: This is due to presence of dissolved bicarbonates of calcium and magnesium. By boiling water, temporary hardness can be removed. Ca(HCO3 )2 Heat  →C CaCO O3 ↓ + H 2 O + CO CO 2 ↑ Calcium lcium bicarbonate lciu (soluble in water)

Calcium lcium caarbonate lciu rbonate (insoluble in water)

Mg( HCO HCO3 )2 Heat  →M MgCO O3 ↓ + H 2 O + CO CO 2 ↑ Magnesium bicarbonate ( soluble in water)

Magnesiium um carbonate (insoluble in water)

(b) Permanent or Non-Carbonate Hardness: This is due to presence of chlorides and sulphates of calcium and magnesium. Permanent hardness cannot be removed by boiling, therefore, special methods are followed. (ii) Hardness is expressed in terms of CaCO3 We know that hardness of water is due to the presence of number of dissolved salts in water but for comparing the hardness of different samples of water of varying composition, it is necessary to choose a reference standard. For this purpose, hardness of water is expressed in terms of equivalents of calcium carbonate only. The following are the reasons for choosing CaCO3 as a standard for expressing the hardness: (a) CaCO3 is a complete insoluble salt; thus it can be easily precipitated completely during water treatment. (b) Its molecular weight is 100, and equivalent weight is 50, so the calculation becomes easy. Hence whatever amount of dissolved salts is present in water, it is fi rst converted into calcium carbonate equivalents by using the formula: CaCO3 equivalent Equivalent weight of CaCO O3 (550) 0) Weight of hardness producing = × substance in mg/L of hardness Equivalent weight of hardness producing substance W × 550 = E W = Mass of hardness-producing substance in mg/L E = Equivalent weight of hardness-producing substance or Mol. wt. of CaCO3 (100) Wt. of hardness producing × substance in mg //L L Mol. wt. ooff the he substance W × 100 = M

Hardness (CaCO3 equivalent) =

Salts responsible for hardness is given in Table 1.1 and method to calculate the hardness.

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Water Technology

1.5

Table 1.1 Calculation of CaCO3 equivalent Dissolved salt/ compound/ ion

Molar mass

Chemical equivalent

Conversion factor for calculation of CaCO3 equivalent

Ca(HCO3)2 Mg(HCO3)2 CaCO3 MgCO3 CaSO4 CaCl2 MgSO4 MgCl2 FeSO4·7H2O CO2 Mg(NO3)2

162 146 100 84 136 111 120 95 278 44 148

81 73 50 42 68 55.5 60 47.5 139 22 74

100/162 100/146 100/100 100/84 100/136 100/111 100/120 100/95 100/278 100/44 100/148

HCO3−

61 17 82 342 40 1 24

61 17 82 57 20 1 12

100/122 100/134 100/164 100/114 100/40 100/2 100/24

OH− NaAlO2 Al2(SO4)3 Ca2+ H+ Mg2+

(iii) Units of hardness Various units used for expressing hardness of water are given below: (a) Parts per million (ppm) (b) Milligrams per liter (mg/L) (c) Degree French (°Fr) (d) Degree Clark (°Cl) (1) Parts per million (ppm): It is defined as the number of parts of calcium carbonate equivalent hardness present per 106 parts of water. 1 ppm = 1 part of CaCO3 equivalent hardness in 106 parts of H2O (2) Milligrams per liter (mg/L): It is defi ned as the number of milligrams of CaCO3 equivalent hardness present per liter of water. 1 mg/L = 1 mg of CaCO3 equivalent per 106 mL of water = 1 part of CaCO3 equivalent per 106 parts of water = 1 ppm (3) Degree French (°Fr): It is defi ned as the number of parts of CaCO3 equivalent hardness present per 105 parts of water. 1 °Fr = 1 part of CaCO3 equivalent hardness per 105 parts of water. (4) Degree Clarkes (°Cl): It is defi ned as the number of parts of CaCO3 equivalent hardness present per 70,000 parts of water. 1 °Cl = 1 part of CaCO3 equivalent hardness per 70,000 parts of water

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1.6

Engineering Chemistry Relationship between various units of hardness is shown in Table 1.2. Table 1.2 Relationship between various units of hardness ppm mg/L °F °Cl

ppm

mg/L

°C

1 1 10 14.3

0.1 0.1 1 1.43

0.07 0.07 0.7 1

Solved Numerical Problems Based on Hardness of Water (i) A water sample contains 204 mg of CaSO4/L. Calculate the hardness in terms of CaCO3 equivalent. Solution CaSO 4 ≡ CaCO CaCO3 136 g mol −1 ≡ 100 g mol −1 ∴136 mg mg / L of CaSO CaSO 4 ≡ 100 mg / L of CaCO O3 eequivalent 100 1 mgg / L of CaSO 4 ≡ 136 100 204 mgg / L of CaaS SO 4 ≡ × 204 mg mg / L of CaCO CaCO3 eequivalent 136 Hardness of CaSO 4 = 11550 mg mg / L of CaCO O3 eequivalent. (ii) Calculate temporary and permanent hardness of a sample of water, which on analysis is found to contain the following: Ca(HCO3)2 = 16.2 mg/L, Mg(HCO3)2 = 7.3 mg/L MgCl2 = 9.5 mg/L,

CaSO4 = 13.6 mg/L

Solution Hardness in terms of CaCO3 equivalent CaCO3 equivalent Substance Ca(HCO3)2 Mg(HCO3)2 MgCl2 CaSO4

CaCO3 equivalent =

Mass of substance (w)

Equivalent mass of substance (E)

16.2 mg/L 7.3 mg/L 9.5 mg/L 13.6 mg/L

81 73 47.5 68

(W ×E 50 ) hardness 10 mg/L 5 mg/L 10 mg/L 10 mg/L

W × 550 E

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Water Technology

1.7

16.2 × 50 = 10 mg/L 81 7.3 × 50 (ii) Mg(HCO3)2, CaCO3 equivalent = = 5 mg/L 73

For (i) Ca(HCO3)2, CaCO3 equivalent =

(iii) MgCl2, CaCO3 equivalent =

9.5 × 50 = 10 mg/L 47.5

(iv) CaSO4, CaCO3 equivalent =

13.6 × 50 = 10 mg/L 68

Now

Temporary hardness = Hardness due to bicarbonate ion = Ca(HCO3 )2 + Mg(HCO3 )2 = 15 mgg / L Permanent hardness = Hardness due to Cll − , S SO O24 − = MgCl2 + CaSO CaSO 4 = 10 + 10 = 20 mg/L

(iii) A sample of water upon analysis gave the following data: MgCl2 = 0.143°Fr, MgSO4 = 0.572°Fr, CaSO4 = 0.286°Fr, and Ca(HCO3)2 = 2.316°Fr. Calculate the hardness in ppm. Solution Since 1°Fr = 10 ppm W × 550 CaCO3 equivalent = E 1.43 × 50 0.143°Fr of MgCl2 = 1.43 ppm = ppm CaCO3 equivalent = 1.51 ppm 47.5 5.72 × 50 ppm CaCO3 equivalent = 4.77 ppm 60 2.86 × 50 0.286°Fr of CaSO4 = 2.86 ppm = ppm CaCO3 equivalent = 2.11 ppm 68 23.16 × 50 2.316°Fr of Ca(HCO3)2 = 23.16 ppm = ppm CaCO3 equivalent = 14.29 ppm 81 0.572°Fr of MgSO4 = 5.72 ppm =

∴ Total hardness in ppm scale = 22.68 ppm (iv) Calculate hardness in terms of CaCO3 equivalent, if 100 ml of a hard water sample neutralizes exactly 12 ml of 0.12 N HCl by using methyl orange indicator. Solution

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N1V1 = N 2 V2 ( Hard water ) ( HCl HCl) N1 × 100 = 0.12 × 12 12 × 12 N1 = 100 × 100

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1.8

Engineering Chemistry Strength of hardness in terms of a CaCO3 equivalent =

12 × 12 × 50 × 1000 100 × 100

Hardness = 720 ppm (v) Calculate carbonate and non-carbonate hardness of water sample, if analysis of water sample gives CO2 = 22 ppm, HCO3− = 305 ppm, Ca2+ = 80 ppm, Mg2+ = 48 ppm and total solids = 5000ppm. Solution

Mass of substance (W)

Equivalent wt. of substance (E)

CO2 HCO3−

22 ppm 305 ppm

22 61

50 ppm 250 ppm

Ca2+ Mg2+

80 ppm 48 ppm

20 12

200 ppm 200 ppm

Constituent

CaCO3 equivalent =

W × 50 E

Carbonate hardness = Due to bicarbonate of Ca and Mg = 250 ppm Non-carbonate hardness = (Hardness due to permanent Ca2+ and Mg2++) − (Hardness due to HCO3− ion) = (200 + 200) – (250) Non-carbonate hardness = 150 ppm

1.5

DETERMINATION OF HARDNESS

We know that there are two types of hardness of water, i.e., temporary and permanent hardness. Temporary hardness is due to bicarbonate of calcium and magnesium, and permanent hardness is due to chlorides and sulphates of calcium and magnesium. The hardness of water can be determined by complexometric titration by using ethylenediamine tetra acetic acid [EDTA] commonly known as EDTA method. EDTA Method: It is the most important and more accurate method to determine the hardness of water. EDTA has limited solubility in water, Hence, disodium salt of EDTA is used which is soluble in water. Principle: EDTA can from complex with salts (Ca2+ and Mg2+) which are present in hard water. Hence, it is known as complexometric titration. Calcium or magnesium ions present in the water sample with ammonical buffer solution form an unstable wine red colour complex with Eriochrome Black T (EBT) indicator. When it is titrated with EDTA solution the metal ions present in water give a stable deep blue colour (M-EDTA) complex and releases the free indicator. The formula of EDTA is written as HOOCH2C CH2COOH N CH2 CH2 N HOOCH2C CH2COOH Ethylene diamine tetra acetic acid

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Water Technology

1.9

Disodium salt of EDTA is HOOCH2C Na+ −OOCH2C

CH2

CH2COO− Na+ CH2COOH

It is represented as Na2H2Y. It ionizes in aqueous solution to give 2Na+ ion and a strong chelating ion represented as H2Y2−. Na 2 H 2 Y → 2 Na + + H 2 Y 2 − It is a hexadentate ligand, and it forms complexes with bivalent cations (Mg2+, Ca2+, etc.), and these complexes are stable in alkaline medium (pH 8-10). EBT may be represented as: O Na+ −O

OH N=N

S O

EBT NO2

{sodium 4-(1-hydroxy-2-napthylazo)-3-hydroxy-7-nitronapthalene-1-sulphonate} The EBT has two ionisable phenolic hydrogen atoms, and it is represented as Na + H 2 In In − ; indicator EBT gives different colours at different pH values. pH 7.0

pH 12.5

3− H H 2 IInn − HIn In 2 − In pH 5.5 pH 10.0

red

blue

yellowish orannge ge

End point: During titration, the colour of the solution changes from wine red to pure blue. Reactions involved during titration: M 2+

(M2 + = Ca2 + /Mgg2 + )

EDTA + EBT EBT → [[M-EBT] Complleex + → [M-EDTA] Complex + EBT wine red (unstable)

(stable)

pure blue

present in hardwater

(i) The calcium and magnesium ion present in hard water combines with the indicator EBT at pH 9-10 to form less stable wine red complex. M 2+ + H HIn In 2 −

→ MInn − + H +

(EBT, Blue)

((M M 2+ = C Caa 2 + / M Mg22++ )

wine red

(ii) When EDTA is added to the water sample, the free M2+ (metal ions) forms a stable complex of M-EDTA. Ca 2+ + H 2 Y 2 − → CaY 2 − CaY + 22H + (M-EDTA Complex)

(EDTA)

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+ H2 Y

2−

(EDTA)

→

MgY 2 −

(M-EDTA Complex)

+ 2H +

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1.10

Engineering Chemistry

(iii) At the end point, when all free metal ions get complexed with EDTA, then further addition of EDTA sets free the metal ion from metal indicator complex and forms more stable metal EDTA complex. CaIn − MgIn −

(wine red) (M-EBT)) (M-EBT (less stab abble) able) le)

+ H2 Y 2− → + H2 Y 2− → (EDTA)

CaY C aY 2 − MgY 2 − MgY

+ +

(Metal EDT EDTA complex) (more stable)

HInn 2 − HI HIn 2 −

EBT indicator (blue)

+ H+ + H+

The metal-EDTA complex may be represented as: O

C H2C

N O CH2

C N

CH2

O O

CH2

H2C

CH2

[M-EDTA Complex] M 2+ = C Caa 2 + //Mg2 + Procedure Step I: Standardization of EDTA solution: Rinse and fill the burette with EDTA solution. Pipette out 50 ml of standard hard water (S.H.W)/Standard MgSO4 solution in a conical flask. Add 10–15 ml of buffer solution and two drops of EBT indicator. Titrate the flask solution against the EDTA solution from the burette until the colour changes from wine red to pure blue, it is end point. Repeat the procedure to get two concordant readings. Let the volume of EDTA be consumed as V1 ml. Step II: Determination of total hardness: Titrate 50 ml of unknown water sample with EDTA solution by addition of 10–15 ml of buffer solution and two drops of EBT indicator till the wine red colour changes to pure blue. Let the volume of EDTA be consumed as V2 ml. Step III: Determination of permanent hardness: Take 250 ml of water sample in a 500 ml beaker and boil gently for half an hour. Cool, filter, and wash the precipitate with distilled water, collecting filtrate and washing in a 250 ml measuring flask, and make the volume up to the mark. Now titrate 50 ml of boiled water sample same as in step I. Let the volume of EDTA be consumed as V3 ml. Observations and Calculations 1 ml of standard hard water = 1 mg of CaCO3 Step I: Standardization of EDTA solution: Volume of S.H.W taken for titration = 50 ml Concordant volume of EDTA used = V1 ml

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Now, V1 ml of EDTA = 50 ml of S.H.W = 50 × 1 mg of CaCO3 50 1 ml of EDTA = mg of CaCO3 V1 Step II: Determination of total hardness: Volume of unknown water sample taken for titration = 50 ml Volume of EDTA used = V2 ml Now, 50 50 ml of unknown water sample = V2 ml of EDTA = V2 × mg of CaCO3 V1 1 ml of unknown water sample = 1 L (1000 ml) of unknown water sample = V2 × 1000 ppm V1

i.e., total hardness =

V2 50 mg of CaCO3 × 50 V1 V2 × 1000 mg/L or ppm V1

Step III: Determination of permanent hardness: Volume of hard water sample taken after boiling and filtering = 50 ml Let concordant volume of EDTA used = V3 ml 50 50 ml of boiled water = V3 ml of EDTA = V3 × mg of CaCO3 V1 V3 50 × 1 ml of boiled water = mg of CaCO3 50 V1 1 L (1000 ml) of boiled water = Permanent hardness =

V3 × 1000 mg/L or ppm V1

V3 × 1000 mg/L V1

Hence, Total hardness =

V2 × 1000 ppm V1

Permanent hardness =

V3 × 1000 ppm V1

Temporary hardness = Total hardness – permanent hardness =

V V2 × 1000 − 3 × 1000 V1 V1

 V − V3  =  2  1000 ppm  V1 

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Advantages of EDTA Method (a) EDTA method shows the result with greater accuracy. (b) This method is more convenient in comparison with other methods. (c) Procedure of EDTA method is more rapid. Solved Numerical Problems Based on EDTA Method (vi) The 1 liter standard hard water (SHW) was prepared by dissolving 1.0 gm of pure and dry CaCO3 in liter distilled water. 50 ml of this SHW required 46 ml of EDTA solution while 50ml of the given hard water sample consumed 20 ml of EDTA solution. After boiling, cooling, and filtering, the hard water sample consumed 10 ml of EDTA solution. Determine the total, permanent, and temporary hardness in ppm. Solution Step-I Standardization of EDTA solution 1L (1000 mL) of SHW = 1 gm (1000 mg) of CaCO3 ∴ 1 ml of SHW = 1 mg of CaCO3 46 ml of EDTA solution required = 50 ml of SHW = 50 × 1 mg of CaCO3 50 1 ml of EDTA solution = mg of CaCO3 46 Step-II Determination of total hardness 50 ml of the given hard water sample required = 20 ml EDTA solution 50 = 20 × mg of CaCO3 46 20 50 × mg of CaCO3 46 50 20 1L (1000 ml) of the given hard water sample required = ×1000 mg/L 46 1 ml of the given hard water sample required =

= 434.78 mg/L Step-III Determination of permanent hardness 50 ml of boiled water sample required = 10 ml of EDTA solution 50 = 10 × mg of CaCO3 eq. 46 10 1 ml of boiled water sample required = mg of CaCO3 eq. 46 10 1L (1000 ml) of boiled water sample required = ×1000 mg/L 46 = 217.39 mg/L Total hardness = 434.78 ppm Permanent hardness = 217.39 ppm Temporary hardness = Total hardness − Permanent hardness = 434.78 − 217.39 = 217.39 ppm

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1.13

(vii) Calculate carbonate and non-carbonate hardness of water, if 20 mL of standard hard water which containing 1.5 g CaCO3 per liter, it required 25 mL EDTA solution for end point and 100 mL of water sample required 18 mL EDTA solution, while the same amount of water after boiling required 12 mL EDTA solution. Solution 1000 ml H2O contains 1.5 g of CaCO3 1 ml of SHW = 1.5 mg of CaCO3 Step-I Standardization of EDTA solution 25 ml EDTA required = 20 ml of SHW = 20×1.5 mg of CaCO3 30 1 ml EDTA required = mg of CaCO3 25 Step-II Determination of total hardness 100 ml of hard water sample required = 18 ml EDTA solution 30 = 18 × mg of CaCO3 eq. 25 18 30 × mg of CaCO3 eq. 100 25 18 30 1L (1000 ml) of hard water sample required = × × 1000 mg/L 100 25 1 ml of hard water sample required =

Total hardness = 216 mg/L Step-III Determination of permanent (non-carbonate) hardness 100 ml of boiled water sample required = 12 ml of EDTA 30 100 ml of boiled water sample required = 12 × mg of CaCO3 eq. 25 12 30 × 1 ml of boiled water sample required = mg of CaCO3 eq. 100 25 12 30 1L (1000 mL) of boiled water sample required = × × 1000 mg/L 100 25 = 144 mg/L Total hardness = 216 ppm Non-carbonate hardness = 144 ppm Carbonate hardness = Total hardness – Non-carbonate hardness = 216 – 144 Carbonate hardness = 72 ppm (viii) Calculate the amount of lime and soda required for the softening of 15000 liters of water, which is analyzed as follows: Temporary hardness = 25 ppm Permanent hardness = 20 ppm Permanent Mg hardness = 15 ppm 74 Solution Lime requirement = (Temp. hardness + perm. Mg hardness) × Volume of water 100 74 (25+15) × 15000 = 444000 mg = 444 g = 100

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Engineering Chemistry 106 (Perm. hardness) × volume of water 100 106 × 20 × 15000 = 318 g = 100

Soda requirement =

∴ Lime requirement = 444 g Soda requirement = 318 g (ix) Calculate the quantity of lime and soda required to soften 20,000 liters of water containing the following salts: CaCO3 = 10.0 mg/L MgCO3 = 8.4 mg/L CaCl2 = 11.1 mg/L MgSO4 = 6.0 mg/L SiO2 = 1.2 mg/L assuming the purity of lime as 90% and soda as 95%. Solution Conversion to CaCO3 equivalent

Constituent

Amount of substance (W)

CaCO3 MgCO3 CaCl2 MgSO4 SiO2

10 mg/L 8.4 mg/L 11.1 mg/L 6.0 mg/L 1.2 mg/L

Lime requirement = =

Equivalent wt. of substance (E) 50 42 55.5 60

CaCO3 equivalent =

W × 50 E

10 mg/L 10 mg/L 10 mg/L 5 mg/L Does not impart hardness

74 (Temp. Ca2+ + 2 × Temp. Mg2+ + Perm. Mg2++) × Volume of water × 100 purity factor 74 100 [10 + 2 × 10 + 5] × 20, 000 × 100 90

Lime requirement = 0.5755 kg 106 [Perm.(Ca2+ + Mg2++)] × Volume of water × purity factor 100 106 100 = [10 + 5] × 220, 000 × 100 95

Soda requirement =

= 0.3347 kg (x) Analysis of water gave the following results H2SO4 = 196 mg/L, MgSO4 = 24 mg/L, CaSO4 = 272 mg/L., and NaCl = 25 mg/L. Water is to be supplied to the town with the population of one lakh only. The daily consumption of water is 50 liter per head. Calculate the cost of lime and soda required for the softening of the hard water for the town for April 2008, if the cost of lime is Rs. 5 per kg and cost of soda is Rs.8 per kg.

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Solution Conversion to CaCO3 equivalents Amount of the substance (W)

Constituent H2SO4 MgSO4 CaSO4 NaCl

196 mg/L 24 mg/L 272 mg/L 25 mg/L

Lime requirement =

Equivalent weight of the substance (E) 49.5 60 68

CaCO3 equivalent =

W × 50 E

200 mg/L 20 mg/L 200 mg/L Does not produce hardness

74 (H SO + MgSO4 as CaCO3 equivalent) × volume of water 100 2 4

50 liter 1kg = 74 (200 + 20) × × 1,00,000 × 6 100 head 10 mg 1kg = 74 (220) × 50 × 105 × 100 106 mg = 814 kg For April 2008, total lime requirement = 814 × 30 = 24420 kg Given cost of lime = Rs. 5/kg Rs 5 Total cost of lime = 24420 kg × kg = 1,22,100 Rs. Similarly, Soda requirement per day 1kg 106 (H 2SO 4 + MgSO4 + CaCO3 ) mg/L × 50 × 105 × 6 100 10 mg 1kg 106 5 = (200 + 2200 + 200) × 50 × 10 × 6 100 10 mg =

= 2226 kg For April 2008 (30 days), total soda requirement = 2226 × 30 = 66780 kg Given cost of soda = Rs. 8.00/kg 8 ∴ Total cost of soda = 66780 × kg = Rs. 5,34,240

1.6

DISSOLVED OXYGEN (DO)

Amount of oxygen dissolved in water (mg/L) is known as dissolved oxygen. At ambient conditions of temperature and pressure, the solubility of oxygen is about 8 mg/L. The amount of dissolved oxygen measures the biological activity of the water bodies, and this is most essential for sustaining aquatic life. Estimation

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Engineering Chemistry

of DO content in a particular water body is of important significance of environmental as well as the industrial point of view. This serves as an indicator of the extent of pollution of water by oxidizable and organic impurities. Further, DO is also responsible for corrosion of boilers and requires to be eliminated. The Winkler test is used to determine the concentration of DO in a water sample. Here the water sample is treated with a mixture of manganese sulphate and alkaline potassium iodide. Initially formed manganous hydroxide precipitate traps the dissolved oxygen and oxidizes manganous ion (Mn+2) to a brown-coloured precipitate of manganic oxide (MnO(OH)2). Mn 2 + + 2OH 2OH − → M Mn(OH)2 1 Mn(OH)2 + 2 O2 → M MnO.(OH) nO.(OH)2 Brown ppt.

The formed manganic oxide precipitate is allowed to settle down for a few minutes and then 2 to 3ml of concentrated H2SO4 is added to dissolved the precipitate. The liberated iodine is proportional to the dissolved oxygen content of water sample. This is estimated by titrating a standard sodium thiosulphate solution and using a starch solution as an indicator. Mn(OH)2 + 4H + + 2I − → Mn 2 + + I 2 + 33H H2O I 2 + 2S2 O32 − → S4 O62 − + 22I − From the above equation, we can fi nd that 1 mole of O2 → 2 moles of MnO(OH)2 → 1 mole of I 2 Therefore, after determining the number of moles of iodine produced, we can calculate the number of moles of oxygen molecule present in the water sample. The oxygen content is usually presented as mg dm−3. The solubility of oxygen in water at ambient conditions of temperature, and pressure is about 8 mg/L.

1.7 DETERMINATION OF CHLORIDES IN WATER Acidity is the ability of water to react with bases and certain metals. (or) An acid is a substance which can act as a proton donor (or) Quantitative capacity of water to neutralise the base. Chlorides are present in water as salts of calcium, magnesium, sodium and potassium [NaCl, CaCl2, KCl, MgCl2]. The salty taste of water is due to NaCl present in it. Chlorides are not considered harmful if their concentration is less than of 250 mg/L. Other salts such as MgCl2 in water undergo hydrolysis and cause problem in boilers. Principle When potassium chromate is added as an indicator to the water sample, it dissolves in water and the chromate ions give yellow colour to the sample. Sodium chloride is present in the dissolved state in the given sample of water. When this is titrated against silver nitrate, the silver ions react fi rst with the chloride ions present in the sample and form silver chloride precipitate and sodium nitrate. AgNO3 + NaCl → AgCl↓ + NaNO3 When all the chloride ions in the sample are precipitated, the excess silver nitrate present reacts with potassium chromate and forms a pale red precipitate of silver chromate. 2AgNO3 + K 2CrO4 → Ag2CrO4 + 2KNO3

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The appearance of the pale red colour indicates that all chloride ions have been precipitated and indicates the end point of titration. From the titre values, the amount of chloride and salt present in the sample is calculated. Indicator: Potassium chromate End point: Yellow to brick red Procedure Take a 50ml burette and wash it with tap water and distilled water and then rinse it with 0.005 N silver nitrate solution. Fill the burette with the 0.005 N silver nitrate solution and note down the initial reading. Pipette out 10 ml of the given water sample with a clean 10 ml pipette into a clean and dry conical flask. Add two to three drops of potassium chromate as the indicator. The solution in the conical flask turns to yellow colour. Titrate this solution against the 0.005 N silver nitrate solution taken in the burette. The appearance of a brick red colour is the end point of titration. Note down the fi nal burette reading. Repeat the titration until consecutive concordant values are obtained. From the titre values, calculate the amount of chloride and salt present in the given water sample using the given formulae. Calculations Pipette solution (water sample) Volume of given water sample (V V1) = 10 ml Normality of given water sample ((N N1) = ?

Burette solution (AgNO3) Volume of silver nitrate solution (V V2) = _____ ml Normality of silver nitrate solution ((N N2) = 0.005 N

N1V1 = N2V2 Normality of given water sample N1 =

N2V2 V1

Amount of chloride present in given sample A = Normality of sample × equivalent weight of chloride = N1 × 35.45 = _____ g/L Amount of salt present in given sample A = Normality of the sample × equivalent weight of chloride salt = N1 × 58.45 = _____ g/L.

1.8

DETERMINATION OF ACIDITY IN WATER

Dissolved carbon dioxide (CO2) in water contributes to the acidity of water by formation of carbonic acid. Water used for drinking purpose should not contain mineral acidity. Highly acidic water, i.e., having low pH affects aquatic life. Principle Acidity of water depends on the end point of indicator used. Hydrolysis or dissociation of acids release H+ ions which react with standard alkali (NaOH) during titrations. The colour change of phenolphthalein indicator indicates neutralization of carbonic acid present in water sample.

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Engineering Chemistry

Indicators (i) Phenolphthalein (ii) Methyl orange End Point In case of methyl orange: Orange to yellow In case of phenolphthalein: Appearance of pink colour Procedure Take a 50 ml burette and wash it with tap water and distilled water and then rinse it with sodium hydroxide solution. Fill the burette with 0.02 N sodium hydroxide solution and note down the initial reading. Take 100 ml of water sample into a conical flask. Add 4 to 5 drops of methyl orange indicator and colour changes to orange. Titrate the water sample against sodium hydroxide solution until the colour changes to yellow and note down the volume consumed as A ml. To the same solution, add 3to 4 drops of phenolphthalein indicator and continue the titration until the appearance of pink colour. Note down the volume of sodium hydroxide consumed as B ml. Repeat the titration to get consecutive concordant readings. Calculations Water sample Normality of water ((N N1) = ? Volume of water (V V1) = 100 ml

NaOH solution Normality of NaOH solution ((N N2) = 0.02 N Volume of NaOH used for methyl orange = A ml Volume of NaOH used for phenolphthalein = B ml Volume of NaOH (V V2) = A + B

N1V1 = N2V2

N1 × 100 = 0.02 × ( A + B) N1 =

0.02 02 × ( A + B ) 100

Acidity of water = Normality × Eq. wt. of CaCO3 = Acidity =

1.9

0.02 02 × ( A + B) × 50 ×1000 × mg/L 100 Volume of NaOH H × N × 550 × 1000 mg/L Volume of water sample

ALKALINITY OF WATER

The capacity of water for neutralizing an acid solution is known as alkalinity of water (or) the capacity of a water to accept protons is known as alkalinity of water.

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The alkalinity of water is mainly dependent on the following factors: (i) Due to the presence of ions like HCO O3− , HSi HSiO H SiO3− , S SiO SiO32 − , etc. in water. (ii) The presence of weak organic acid salts. Because they consume or have a tendency to take up N+ ions, hence concentration of OH– ions in water increases. Classification of Alkalinity Depending on the ions present, alkalinity of water is broadly classified as (i) Hydroxide (OH–) 2− (ii) Carbonate (CO3 ) and − (iii) Bicarbonate (HCO3 ) alkalinity

The alkalinity of a water sample may be due to (i) OH− (ii) CO32− (iii) HCO3− (iv) OH− and CO32− (v) CO32− and HCO3− But there is no possibility with OH − andd HCO HCO3− , because they combine with each other to form carbonate. OH − + HCO HCO3− → C CO O32 − + H 2 O Units Alkalinity and hardness are expressed in terms of CaCO3 equivalents, ppm, mg/L, etc. (i) Alkalinity < total hardness Carbonate hardness in ppm = Alkalinity in ppm (ii) Alkalinity ≥ total hardness Carbonate hardness in ppm = Total hardness in ppm Non-carbonate hardness = Total hardness – Carbonate hardness Determination The type and extent of alkalinity of a water sample can be easily determined by volumetric method. Aknown volume of water sample is titrated against standard sulphuric acid by using phenolphthalein indicator. The end point is disappearance of pink colour. Further the titrated water sample is titrated against the same standard sulphuric acid by using methyl orange indicator. The end point is appearance of red colour and the volume of H2SO4 consumed is noted.

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 (i) OH H − + H+ → H2O   2− − + − P (ii) CO3 + H → H HCO CO3  M − + (iii) HCO O3 + H → H 2 O + CO2  The volume of the standard acid used up to phenolphthalein end point P marks the completion of reactions (i) and (ii), whereas the total volume of the standard acid used from the beginning up to methyl orange end point M corresponds to the completion of reactions (i), (ii) and (iii). Solved Numerical Problems Based on Alkalinity of Water N H SO for neutralization to phenolphthalein end 50 2 4 point. Another 16 mL of same acid was needed for further titration to methyl orange end point. Determine the type and amount of alkalinity in terms of CaCO3 equivalent. Solution Volume of water sample for titration = 100 mL Volume used to phenolphthalein end point (A) = 4 mL Volume used to methyl orange end point (B) = 16 mL Total volume used to methyl orange end point (A + B) = 20 mL Phenolphthalein alkalinity (in terms of CaCO3 equivalent)

(xi) 100 mL of a water sample required 4 mL of

N1V1 = N 2 V2 ( Water ) ( Acid ) N ×4 50 4 N1 = 50 × 100

N1 × 100 =

Strengthh = N1 × Eq. wt of CaCO3 4 × 50 50 × 100 4 Phenolphthalein alkaliinity nity (P) = × 50 ×1000 × ppm 50 × 100 = 40 ppm Similarly, for methyl orange alkalinity, N3V3 = N4V4 Water

Acid

N × 20 50 N 20 N3 = × 50 100

N3 × 100 =

Strength = N3 × Eq. wt. of CaCO3 N 20 = × × 50 50 100

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Water Technology Methyl orange alkalinity (M) =

1.21

1 20 × × 50 × 1000 ppm 50 100

M = 200 ppm 1 Hence, P< M 2 1 P(40) < M(100) 2 So CO32− and HCO3− ions are present. Now, alkalinity due to CO32− ions = 2P = 2 × 40 ppm = 80 ppm alkalinity due to HCO3− ions = M – 2P = 200 – 80 = 120 ppm (xii) Calculate the alkalinity in CaCO3 equivalents, if 100 mL of a water sample on titration with 0.03 N HCl by using phenolphthalein indicator and end point is at 7.5 mL acid. Another water sample of same volume require 15 mL of same concentration acid by using methyl orange indicator to obtain complete neutralization. Solution Volume of water sample = 100 mL For Phenolphthalein alkalinity N1V1 = N 2 V2

( Water )

( Acid )

N1 × 100 = 0.03 × 7.5 0.03 × 7.5 N1 = 100 Strengthh = N1 × Eq. wt. of CaCO3 0.03 × 7.5 = × 50 100 0.03 03 × 7.5 Phenolphthalein alkalinity (P) = × 50 ×1000 × ppm 100 = 112.5 ppm For Methyl orange alkalinity, N3V3 = N4V4 (Water)

(Acid)

N3 × 100 = 0.03 × 15 N3 =

0.03 03 × 115 100

Strength = N3 × Eq. wt of CaCO3 =

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0.03 03 × 115 × 50 100

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1.22

Engineering Chemistry Methyl orange alkalinity (M) =

0.03 03 × 115 × 50 × 100 100

M = 225 ppm 1 P= M 2

Hence

1 225  P(122.5) = M  2  2  So alkalinity due to CO32− ions, Alkalinity due to CO32− ions = 2P or M = 225 ppm N H2SO4 for phenolphthalein end point and 50 another 5 ml for methyl orange indicator, i.e., complete neutralization. Describe the type and amount of alkalinity. Solution For Phenolphthalein alkalinity,

(xiii) 50 ml of alkaline water sample required 20 ml of

N1V1 = N2V2 (Water)

(Acid)

1 × 20 50 1 20 × N1 = 50 50

N1 × 50 =

Strength = N1 × Eq. wt. of CaCO3 20 × 50 = 50 × 50 ∴ Phenolphthalein alkalinity (P) =

20 × 50 × 1000 ppm 50 × 50

= 400 ppm

For Methyl orange alkalinity,

N3V3 = N4V4 (Water)

(Acid)

N3 × 50 =

1 × 25 50

N3 =

25 50 × 50

Strength = N3 × Eq. wt. of CaCO3 =

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25 × 50 50 × 50

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Water Technology ∴ Methyl orange alkalinity (M) = Hence,

1 P> M 2

1.23

25 × 50 × 1000 50 × 50

= 500 ppm

1 P(400) > M  500  2  2  So OH– and CO32− ions are present. Now, alkalinity due to OH– = 2P – M = 2(400) – 500 = 300 ppm 2− alkalinity due to CO3 = 2(M – P) = 2(500 – 400) = 200 ppm

1.10

DISADVANTAGES OF HARD WATER

Hard water contains large amounts of bicarbonates, sulphates, and chlorides of calcium and magnesium salts. It causes number of problems in domestic use, industrial use, and in boilers. (i) Problems in Domestic Use: (a) Cooking: Pulses and other vegetables do not cook well in hard water. Tea, coffee and other drinks prepared with hard water gives an unpleasant taste. (b) Drinking: Hard water causes bad effect on digestive system. Due to formation of calcium oxalates, stones are formed in kidneys. (c) Wastage of soap: Washing: Hard water does not give much lather with soap, as most of the soap is consumed for removing calcium and magnesium salts present in water. Bathing: It produces sticky scum on both tub and body. (d) Damaging clothes: The Ca2+ and Mg2+ ions present in hard water combine with soap to form insoluble compounds, which sticks to the clothes. This is difficult to remove, and hence damages the clothes. (e) Wastage of fuel: Much fuel is consumed for boiling hard water in kettles because salts form an incrustation inside the kettle due to formation of carbonates and hydroxides of calcium and magnesium. After prolongedusage, kettle also gets damaged due to scale formation. (ii) Problems in Industrial Use: (a) Textile industry: Water is used in textile industry for cleaning, washing, and whitening of yarn. For such purposes, soap is required; if hard water is used, more amount of soap is wasted. (b) Paper industry: The water that is used in paper industry should be free from hardness, suspended particles, iron, etc. Because hardness increases the ash contents of paper, suspended particles produce cracking tendency of paper, and the salt of iron decreases the brightness of paper.

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Engineering Chemistry

(c) Iron industry: Hard water makes the iron of low quality. It corrodes the iron and its alloys. (d) Dyeing industry: The water that is used for dyeing purpose should be free from hardness, because salts of calcium and magnesium spoil the desired shade. (e) Sugar industry: Water should be free from hardness-suspended particles as well as pathogenic microorganisms because hard water causes difficulties in the crystallization of sugar from molasses. (iii) Problems in Boilers, Use: For the generation of steam a huge quantity of water is used in boilers and is known as boiler feed water. If water used for boilers is hard, it may create number of problems like caustic embrittlement, corrosion, scale and sludge formation, priming and foaming, etc. This is very dangerous because at high pressure the same causes explosions. Hence water which is used in boilers should be softened and should be pure before feeding into the boilers. Boiler-feed water should satisfy the following requirements: (a) Hardness < 0.5 ppm (b) Caustic alkalinity = 0.15 – 0.45 ppm (c) Soda alkalinity < 1 ppm (d) Excess soda ash < 0.55 ppm

1.11

QUALITY OF WATER FOR DOMESTIC USE

The potable water or drinking water should satisfy the following essential requirements: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii)

1.12

Water should be clear, clean, colourless, and odourless. Total dissolved solids (TDS) should be less than 500 ppm. It should be free from hardness, suspended particles, and pathogenic bacteria. Turbidity should be less than 10 ppm. Its pH should be about 7–8. It should be free from harmful, dissolved solids like arsenic, manganese, chromium, lead, etc. It should be free from harmful gases like H2S, SO2, etc. It should be neither too hard nor too soft. The recommended hardness is about 300 mg/L as CaCO3 equivalent. Its alkalinity should not exceed 600 mg/L. It should have an agreeable taste. Fluoride should be less than 3 ppm. Chloride and sulphate must be less than 250 ppm.

TREATMENT OF WATER FOR DOMESTIC USE

Purification of water for potable use involves mainly the following steps: (i) (ii) (iii) (iv) (v)

Screening: Removes the floating materials like leaves. Sedimentation: Removes suspended impurities like sand, clay, etc. Coagulation: Removes finely divided suspended particles. Filteration: Removes colloidal impurities and large organisms. Disinfection: Kills the bacteria.

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(i) Screening: The raw water is allowed to pass through screens having large number of holes, where floating impurities like rags, paper, leaves, etc., are held by them, and water is passed through the holes. (ii) Sedimentation: Sedimentation is a process for retention of water for certain period in a deep tank ( ~ 5 meter) or to flow quietly at low velocities. Most of the suspended particles settle down due to the force of gravity. This process takes two to eight hours. This process removes 70%–75% of suspended impurities. (iii) Coagulation: Coagulation is the process by which the fine, suspended, and colloidal impurities are removed from the water by the addition of suitable chemicals (coagulants). The finely divided suspended inorganic matters do not settle down so easily, so these smaller particles are converted into larger ones, which have higher settling velocities. The commonly used coagulants are the salts of iron and aluminium, e.g., alum (K 2SO4 Al2(SO4)3 . 24H2O), ferrous sulphate (FeSO4 . 7H2O), sodium aluminate (NaAlO2), etc. These coagulants react with alkaline salts and form a thick gelatinous precipitate known as Flock. Flock has the property to attract finely suspended particles and form big flock, which settles down rapidly. This process is called flocculation. A few commonly used coagulants and their reactions are as follows: (a) Alum (Al2(SO4)3 K 2SO4 . 24H2O) Al 2 (SO (SO 4 )3 + 3Ca(HCO 3Ca(HCO3 )2 → 2Al(OH)3 ↓ + 3CaSO O 4 + 66CO CO2 (Flocculant)

Al 2 (S SO O 4 )3 + Mg(HC Mg(HCO M g(HCO O3 )2 → 2Al(OH)3 ↓ + 3MgSO 3MgSO 4 + 66CO2 (Flocculant)

(b) Sodium aluminate (NaAlO2) NaAlO O2 + 2H 2H 2 O → Al(OH)3 ↓ + NaOH (Floocculant cculant)

The NaOH produced precipitate of Mg salts as Mg(OH)2 MgSO 4 + 2NaOH 2NaOH → M Mg(OH) g(OH)2 + Na 2SO 4 (c) Ferrous sulphate (FeSO4 . 7H2O) FeSO 4 + Mg(HCO3 )2 → Fe(OH)2 ↓ + MgSO 4 + H 2 O + CO2 FeSO 4 + Ca(HCO Ca(HCO3 )2 → Fe(OH H))2 ↓ + CaSO CaSO 4 + H 2 O + CO2 4Fe(OH) 2 + 2H 2 O + O2 → 4Fe(OH)3 ↓ (Flocculant)

The precipitates obtained by using suitable coagulants in water get settled down during sedimentation. (iv) Filteration: Almost all suspended impurities are removed through filteration process. During filteration, all types of insoluble colloidal and bacterial impurities are also removed by passing water through a bed of proper-sized material. Two types of filters are commonly used in domestic water treatment. (a) Gravity sand filter (b) Pressure filter

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Engineering Chemistry

sedimented water (Inlet)

Fine sand

30-60 cm

Coarse sand Gravels

30-60 cm

Under drain channel

Filtered water (Outlet)

Figure 1.3 Sand filter (a) Gravity sand filter: It consists of a large, shallow, rectangular tank made of concrete and a process medium, known as filter medium, which retains solid particles but allows the passage of water as shown in Figure 1.3. It consists of three layers. The upper layer consists of fi ne sand (about 50 cm thick) and is a thick layer. The middle layer consists of coarse sand (about 20 cm thick), and the bottom layer consists of gravels (about 30 cm thick). It is provided with an inlet for sedimented water and an under drain channel at the bottom for the exit of filtered water. Sedimented water enters the sand filter from the top and is uniformly distributed over the fi ne sand layer. As the water percolates through the sand bed, fi nely suspended particles and most of the germs and bacteria are retained by the top layer. Clear, filtered water is collected in the under drain channel, from where it is drawn out. The rate of filteration becomes slow after some time due to clogging of pores of the top sand layer by the impurities retained in the pores. Therefore, the portion of the top fine sand layer is scrapped off and replaced by a new sand layer. The filter is put to use again. (b) Pressure filter: It consist of a cylindrical as shown in Figure 1.4, vertical steel tank containing three layers of filtering media, one above the other. (1) Pebbles layer (10–35 mm grain size) (2) Coarse sand layer (5–7 mm grain size) (3) Fine sand layer (1–2 mm grain size) Impure, sedimented water is mixed with a small amount of alum solution, and then water is forced through filter bed under pressure. Alum forms the slimy layer on the filter bed, and this helps in the removal of colloidal and bacteriological impurities. The function of deflector plate, which is provided at the top, is to distribute the slimy layer of alum uniformly over the top of the filter bed. Filtered water, as it comes out from the bottom of filter, is under pressure and can thus be pumped directly. These filters are widely used for industrial purposes. (v) Disinfection/Sterilization: Sterilization of water means complete destruction of all living microorganisms (bacteria, virus, etc.) present in water. We know that water after passing through different processes such as sedimentation, coagulation, and filteration processes still contains a small percentage of pathogenic bacteria. Therefore, it is necessary to remove these bacteria and microorganisms from water. The chemicals used for sterilization are known as sterilizers or disinfectants.

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1.27

Deflector plates

Wash water (outlet)

Slimy layer of Al(OH)3 Fine sand Coarse sand Pebbles

Compressed air inlet

Wash water inlet

Filtered water outlet

Figure 1.4 Vertical pressure filter Several methods have been adopted for sterilization of water. Some of them are given below: (a) (b) (c) (d) (e)

Boiling method Chlorination method Ozonolysis method UV-rays method Membrane technology method

(a) Boiling method: Water for domestic purposes on a smaller scale may be sterilized by simple boiling method. In this method, water is boiled for about 20–30 min. This method kills the harmful disease-causing bacteria and germs. But this method is useful only for household purposes because this process is very much expensive for municipal supply of water, and in addition, a large quantity of fuel is required to boil water on a large scale. It does not provide any protection for further contamination of water. (b) Chlorination method: It is the most important method for sterilization of water. Chlorination is done by the following methods: (1) By using chlorine gas or concentrated aqueous solution. (2) By using bleaching powder. (3) By using chloramine. (1) By using chlorine gas or concentrated aqueous solution Chlorine is a powerful germicide and most commonly used disinfectant. Chlorine used for this purpose can be used directly as a gas or as chlorine water. It reacts with water to form hypochlorous acid and nascent oxygen, both of which are powerful germicides.

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Engineering Chemistry Cl 2 + H 2 O → HCl + HOCl Hypochlorous aaci cid

HOCll → HCl HCl + [O]

(nasceent nt oxygen)

Germss + [O] [O] → Germs are oxidised HOCll + bacteria bacteria → Deacttivated ivated bacterias Apparatus: The apparatus used for disinfection by chlorine is known as chlorinator (Figure 1.5). It is a large tower containing number of baffle plates. From the top of the tower, proper dose of chlorine and water is introduced. They get thoroughly mixed during their passage through the tower, and treated water is taken out from the bottom. Advantages (i) (ii) (iii) (iv) (v)

It is cheap and is an easily available disinfectant. At a low concentration, it is very effective bactericide. It can be used at high and low temperatures. It is stable and does not deteriorate on keeping. Chlorine residue can be maintained in treated water, which provides additional safety for preventing regrowth of bacteria.

Disadvantages (i) Excess of chlorine produces an unpleasant taste and odour in water. (ii) It is less effective at higher pH value but more effective at lower pH value (below pH 6.5).

Concentrated chlorine solution

Raw water inlet

Baffle plates

High tower

Sterilized water outlet

Figure 1.5 Chlorinator

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(2) By using bleaching powder (CaOCl2) Bleaching powder is a strong oxidizing agent and is having 30 per cent available chlorine. When water is treated with bleaching powder, hypochlorous acid is formed. Itreleases nascent oxygen and the nascent oxygen thus released deactivates the enzymes of microorganisms; due to this, metabolic activities will stop and the microorganisms get killed. CaoCll 2 + H 2 O  →C Caa(OH OH)2 + C Cll 2 ( Bleaching Bleaching powder powder )

Cl 2 + H 2 O    → HCl HCl + H HOCl ( hypochlorous acid acid )

HOCll  →H HCl Cl + [O] Nascent oxygen

Germss + [O]    → Deactivate the enzyme ↓ Stop metabolic activities of miicroorganism croorganisms ↓ Kill About 1 kg of bleaching powder is sufficient for 1000 kilolitres of water, but allow the water to stand for several hours. Disadvantages (i) Excess of bleaching powder creates bad taste and odour to water. (ii) It introduces calcium hardness in water due to the formation of Ca(OH)2. (iii) It is unstable, so its storage is difficult. (3) By using chloramines (NH2Cl) By mixing of chlorine and ammonia in 2:1 ratio, chloramine is formed. Cl 2 + NH 3    → NH 2 Cl + HCl ( chloramine chloramine)

Whenever water is treated with chloramine, hypochlorous acid is formed and with release of hypochlorous acid it provides greater safeguard from recontamination. ClNH 2 + H 2 O    → HOCl + NH 3 Hypochlorous acid

HOCll  →H HCl Cl + [O] Germss + [O]    → Kills the germs So, HOCl + germs → germs are killed. Advantages (i) Excess dose of ClNH2 does not create bad odour and taste in water. (ii) It provides a greater lasting effect than chlorine.

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Engineering Chemistry Waste gases

Impure water (inlet)

Gravel bed

Curved partition having perforation

Ozone gas (inlet) Sterilized water (outlet)

Figure 1.6 Ozone sterilizer (c) Ozonolysis method: Ozone is used for this method. Ozone is a highly unstable and excellent disinfectant. It breaks down and gives nascent oxygen. O3 → O2 + [O]

(Nascent oxygen)

The nascent oxygen is very powerful oxidizing agent, which kills all the bacteria and germs present in water. Apparatus: The reaction of ozone and water is carried out in ozone sterilizer (Figure 1.6). During the treatment of water, water is allowed to enter from top to bottom, and ozone is allowed to enter from bottom to top, which kills the germs when they come in contact with each other. Sterilized water is collected at the bottom of the tank. The contact time for ozone and water is about 10–15 minutes. Advantages (1) It removes colour and odour from water. (2) It improves the taste of water. (3) The excess dose of ozone is not harmful, because it releases O2 on decomposition. (d) UV-rays method: When water is exposed to UV-rays from electric mercury lamp that is immersed in water, most of the pathogenic bacteria are destroyed. This method is widely used for the disinfection of swimming pool water. Advantages (1) It does not require any chemicals. (2) It has not any bad effect during treatment. (3) It does not produce any odour in water.

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Disadvantages (1) It is very expensive, so it is not widely used on a large scale. (e) Membrane technology method: Disinfection by this method is generally used for drinking water. In this method, water purifies most of the contaminant ions, molecules, and small particles including viruses and bacteria by passing them through a membrane having uniform microscopic-size pores. Membrane processes include microfilteration, ultrafilteration, nanofilteration, and reverse osmosis. In all these methods, water is forced through membranes made of synthetic polymers, cellulose acetate, or even ceramics by the application of high pressure in the range of 10 to 50 atm. pressure. Microfilteration and ultrafilteraton membranes with pores of 0.002 to 10 μm in diameter can filter off most bacteria and colloidal particles but not viruses and ions. Nanofilteration soften water by removing hardness causing metal ions, and reverse osmosis is used for desalination of sea water.

1.13

BREAK-POINT CHLORINATION

Chlorination of water is done carefully in a controlled manner with the dip or break is called breakpoint chlorination. Added chlorine consumed for different reactions such as (i) Oxidation of reducing substance (ii) Chlorination of organic substance (iii) Oxidation of ammonia and disinfection of bacteria With this method not only living organisms but also organic impurities and free NH3 present in water are destroyed. The point at which free residual chlorine begins to appear is called break-point chlorination. It is also known as free residual chlorination. Break-point chlorination shows four stages of sterilization as shown in Figure 1.7:

Residual chlorine

Stage-III Destruction of chloro Stage-I organic and Stage-II chloramOxidation Formating of B ine of reducing chloro organic compounds and compounds chloramine by compounds chlorine

Stage-IV Free residual chlorine D

C Break point

Combined chlorine O

Free chlorine

A Applied chlorine dose (mg/L)

Figure 1.7 Break-point chlorination curve

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Stage I: Initially, with the lower dosage of Cl2, there is no free residual chlorine since all the added Cl2 gets consumed in complete oxidation of reducing substances present in water. Stage II: As the amount of Cl2 dose is increased, the amount of residual Cl2 also shows steady increase. This stage corresponds to the formation of chloro-organic and chloramines compounds without undergoing oxidation. Stage III: As more amount of Cl2 is applied, the amount of free residual chlorine also decreases, due to oxidation of chloro-organic and chloramines. When the oxidation destruction is complete, it reaches a minima. Stage IV: After minima, the added Cl2 is not used in any reaction. Thus, the residual Cl2 keeps on increasing in direct proportion to added Cl2. The point ‘C’ is called break point. It is a point at which free residual chlorine begins to appear. Thus, break-point chlorination helps in eliminating disagreeable odour and bad taste in water. Advantages (i) (ii) (iii) (iv)

It prevents the presence of excess chlorine, which may impart bad odour and taste to water. It ensures complete destruction of disease-producing bacteria. It prevents the development of any weeds in water. It helps to calculate the sufficient amount of chlorine for adding in water.

Dechlorination: Excess of chlorine after the break-point chlorination gives unpleasant taste and odour in water. The excess of Cl2 may be removed by filtering the treated water over activated carbon. Over chlorination may also be removed by treating the water with SO2, Na2SO3, and Na2S2O3. SO2 + Cl 2 + 2H 2 O → H 2SO 4 + 2HCl Na 2SO3 + Cl 2 + H 2 O → Na 2SO 4 + 2HCl Na 2S2 O3 + 44C Cl 2 + 5H 5H 2 O → 2NaHSO4 + 8HCl Superchlorination: Superchlorination is the addition of excess amount of chlorine for disinfection of water. It destroys the pathogenic microorganisms as well as organic impurities by oxidation. Prechlorination: Prechlorination is the treatment of water with chlorine before filtration. In this process high chlorine is required to satisfy the chlorine demand of filterable matter. With prechlorination the quality of water is superior because unpleasant tastes and odours due to chlorinated products may be absorbed during filtration. This process is highly expensive. Post-chlorination: Post-chlorination is the treatment of chlorine with water after filtration. In this method treated water may have unpleasant taste and odour, but it is cheaper than prechlorination due to lower chlorine demand.

1.14

BOILERS AND BOILER TROUBLES

In all the industries, boilers are used for generating steam. Boiler-feed water is the water required for generation of steam and with the safety, economy and efficiency concerns it should be of very good quality.

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Depending upon the operating pressures, boilers are classified into low-pressure (10–15kg/cm 2), medium-pressure (15–35 kg/cm 2), high-pressure (50–140 kg/cm 2), very high-pressure (150–225 kg/ cm2) and supercritical boilers (>225kg/cm2). Depends upon the quality of the feed water, so many problems may arise in the boilers. Some of them are listed hereunder. (i) (ii) (iii) (iv)

Scale and sludge formation Priming and foaming or carry over Boiler corrosion Caustic embrittlement

(i) Scale and Sludge formation In the boilers, when water is vaporized to steam gradually the concentration of dissolved salts increases. When the concentration of salts reaches their saturation, they are thrown out in the form of precipitates. Sludge is the soft, slimy and non-adherent layer of precipitate inside the boiler and also called mud. Hard adhering coating of precipitate inside the boiler walls is called scale. Scale and sludge are shown in Figure 1.8. (a) Sludge: Sludge is a soft, loose, and slimy precipitate formed within the boiler sludge. It can easily be scrapped off with a wire brush. It is formed at comparatively colder portions of the boiler and collects in areas of the system, where the flow rate is slow or at bends. Sludges are formed by substances that have greater solubility in hot water than cold water. Composition: The main composition of sludge includes MgCO3, MgCl2, CaCl2, MgSO4,etc. Disadvantages (1) Sludges are poor conductor of heat, so they tend to waste a portion of heat generated. (2) Sludges decrease the efficiency of the boiler. (3) Since sludges settle in areas of poor water circulation such as joints, bends, etc., therefore choking of pipes takes place. Removal of Sludge (1) By taking small precautions, like using of soft water, prevent the formation of sludge (2) Scrapping of sludge with hard brush (3) With frequent ‘blow-down operation’ i.e., replacement of concentrated water with fresh water. Water

∼ ∼∼ ∼ ∼ ∼∼ Loose, soft, slimy ppt (sludge)

Boiler wall Heat

Heat

Hard, adhering coating on inner walls of boiler (scale)

Figure 1.8 Sludge and scale

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Engineering Chemistry (b) Scale: Scale is a hard, adhering and sticky deposit. They stick very fi rmly on the inner walls of the boilers and very difficult to remove even with a chisel and hammer. These are formed with CaCO3, Ca(OH)2, Mg(OH)2, CaSO4, CaSiO3, MgSiO3, etc. Formation of Scales: Due to the following reactions, scales are formed. (1) Decomposition of calcium bicarbonate In low pressure boilers, calcium bicarbonate decomposes and gives calcium carbonate. Ca( HCO HCO3 )2 Heatin g →C CaC aCO O3 ↓ H 2 O + CO CO 2 ↑ ( scale scale)

At high pressure boiler formed CaCO3 is soluble and gives calcium hydroxide, whose solubility decreases with the temperature and deposit as scale. High temperature

CaCO3 + H 2 O    → Ca(OH)2 ↓ + CO2 ↑ ( scale scale)

(2) Deposition of calcium sulphate as a scale With increase of temperature the solubility of calcium sulphate decreases, and consequently gets precipitated as hard scale. This scale is quite adherent and difficult to remove. (3) Hydrolysis of magnesium chloride At high temperatures, the magnesium salts undergo hydrolysis and give magnesium hydroxide. MgCl2 + 2H 2 O  → Mg(OH)2 ↓ + 2HCl ( soft scale scale)

(4) Formation of silicates Minute amounts of silica present in water form and deposit as calcium or magnesium silicates and stick very fi rmly to the inner side of the boiler surface. Disadvantages (i) Fuel wastage: As scales have a low thermal conductivity to provide a continuous supply of heat to water, overheating is done, which results in the wastage of fuel. (ii) Decrease in efficiency: Scales get deposited in the valves and condensers of the boilers, thereby choking them partially. It results in decrease in efficiency of the boiler. (iii) Danger of explosion: Sometimes at high pressure, the scales crack and water suddenly comes in contact with overheated iron plates. This results in the sudden formation of large amount of steam, which may cause explosion. (iv) Lowering of boiler safety: Super heating of boiler makes the boiler material softer and weaker, which causes distortion of boiler tube. Removal (i) For soft scale: Soft scale is loosely adhering, so it can be removed with the help of wire brush or blow-down operation. (ii) For brittle scale: Brittle scale can be removed by giving thermal shocks to the boiler, i.e., heating and cooling suddenly. (iii) For hard and adhering scale: They can be removed by adding chemicals.

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Example: (i) CaCO3 scale can be dissolved by using 5%–10% HCl. (ii) CaSO4 scale can be dissolved by adding EDTA since Ca-EDTA complex is highly soluble in water. H 2 Y 2 − + Ca 2 + → CaY Y 2 − + 2H + (ii) Priming, carry over and foaming (a) Priming: Priming is the carrying of small droplets of water along with steam while boiling the water. This is also known as wet steam. Causes of Priming (1) (2) (3) (4) (5) (6) (7)

Improper design of the boiler Presence of large amount of fi nely divided particles in the boiling water Presence of large amount of dissolved solids Very high level of water in the boiler High steam generation velocities Sudden increase in steamproduction rate Presence of foam on the surface

Precautions to Reducing Priming (1) (2) (3) (4) (5)

Maintaining low level of water Fitting of mechanical steam purifiers Ensuring efficient softening and filtration of boiler-feed water Avoiding of rapid change in steam generation Removal of scale and sludge frequently Carryover: Carrying of suspended and dissolved solids along with wet steam is called carryover.

(b) Foaming: Foaming is the formation of persisted form (or) stable bubbles in the boilers, which do not break easily. This is due the concentration difference of suspended solid between the film and the bulk of water. It is also due to the presence of oil and alkalies in boiler-feed water. Oily substances and alkalies react to form soaps, which reduce the surface tension of water and thus increase the foaming tendency of water. Causes of Foaming The following are the causes of foaming: (1) Due to the presence of oil or grease in water (2) Due to the presence of fi nely divided sludge in water (3) Due to the presence of some chemicals, which reduce the surface tension. Prevention of Foaming (1) By the addition of antifoaming agents such as castor oil or polyamides in low-pressure boilers (2) By using soft and filtered water

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Engineering Chemistry (3) By removing oil from boiler water by adding coagulants like ferrous sulphate or sodium aluminate, etc. Disadvantages of Priming and Foaming (1) Due to the presence of foaming, boiling point of water is increased; hence wastage of fuel occurs. (2) It is very difficult to maintain the constant pressure of steam. (3) Due to the excess formation of foaming, the bubbles entered into the engine along with the steam, which lowers the efficiency of engine. (4) Due to priming and foaming, corrosion takes place in the part of the engine. (5) Due to the presence of foam, water level is not identified.

(iii)

Boiler Corrosion It is the disintegration or decay of boiler material either due to chemical or electrochemical reaction with its environment. Factors Causing the Boiler Corrosion (a) Formation of rust with dissolved oxygen: Dissolved oxygen present in the water attacks the boiler material and easily forms rust. 2Fe + 2H 2H 2 O + O2 → 2Fe(OH)2 ↓

Ferrous hydroxide

4 Fe(OH)2 + O2 → 2[Fe2 O3 ⋅ 2H 2H 2 O] Rust

(b) Due to the presence of dissolved CO2: The source of CO2 in water, either dissolved CO2gasor bicarbonates, on heating gives CO2 and is also responsible for boiler corrosion. Mg(HCO3 )2 ∆ → Mg(OH Mg(OH)2 + 2C CO2 ↑

Ca(HCO3 )2 ∆ →C CaC aCO3 ↓ + H 2 O + C aCO CO2 ↑

Carbon dioxide (CO2) dissolves in water to form a weak carbonic acid. CO2 + H 2 O → H 2 C CO O3 H 2 CO CO3 + Fe Fe → F FeCO FeC eCO3 + H 2 eCO (c) Due to the formation of acids from dissolved salts: Chlorides of some inorganic salts like MgCl2, CaCl2 etc., which present in water can produce hydrochloric acid and can corrode the boilers. MgCl2 + 2H 2 O → Mg(OH)2 ↓ + 2HCl CaCl 2 + 2H 2 O → Ca(O Ca(OH Ca (OH H ) 2 ↓ + 2H HCl The liberated HCl reacts with boiler material in chain-like reaction. Fe + 2HCl HCl → F FeCll 2 + H 2 FeCl 2 + 2H 2 O → Fe(OH)2 + 2HCl (d) Due to the presence of oil: Oil undergoes hydrolysis, releasing free fatty acids leading to corrosion of the boiler.

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Prevention of Boiler Corrosion (a) By removal of DO (1) Preheating: As solubility of gases decreases with increases in temperature, at approximate 65°C, complete DO is removed. (2) Chemical treatment: The DO is removed through the addition of Na2SO3 or Na2S or hydrazine (N2H4). 2 Na 2SO3 + O2 → 2 Na 2SO 4 Sodium Sulphite

Sodium Sulphate

Na 2S + 2O2 → Na 2SO SO 4

Sodium Sulphide

N 2 H 4 + O 2 → N 2 ↑ + 2H 2 O

Hydrazine

Hydrazine is found to be an ideal compound for removing DO because the products are water and nitrogen gas, which do not form hard products, while due to sodium sulphite (Na2S) and sodium sulphide (Na2SO3), there is a formation of sodium sulphate (Na2SO4), which decomposes and gives SO2, and it forms sulphurous acid (H2SO3) in steam condensate. (3) Mechanical deaeration: As shown in Figure 1.9, the water passes through the perforated plates and undergoes deaeration at high temperature and low-pressure dissolved oxygen and carbon dioxide escapes. The solubility of a gas in water is directly proportional to pressure and inversely proportional to temperature, hence the water gets deaeration. Boiler feed water (inlet)

To vacuum pump

Steel jacket

Perforated plate

T Tower

Deaerated water (outlet)

Figure 1.9 Mechanical deaeration of water

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Engineering Chemistry (b) By removal of dissolved carbon dioxide (CO2) (1) Preheating: By increasing the temperature, solubility decreases. (2) Chemical treatment: By adding calculated quantity of ammonia 2 NH 4 OH + CO2 → ( NH 4 )2 CO3 + H 2 O (3) Mechanical deaeration: It removes DO as well as CO2 from feed water. (c) Addition of alkali: Corrosion by acids may be prevented by adding some alkalies from outside so that product acid may be neutralized. (d) By using soft water in the boiler for steam generation.

(iv) Caustic Embrittlement Caustic embrittlement is the special type of boiler corrosion caused by the use of highly alkaline water. With this phenomena boiler material becomes brittle with the accumulation of caustic substances. During the softening of water by lime soda process, usually small amount of free Na2CO3 is present. In high-pressure boilers, sodium carbonate decomposes and gives sodium hydroxide and this makes the boiler water ‘caustic.’ Na 2 CO3 + H 2 O    → 2 NaOH + CO2 ↑ The concentration of NaOH is increased by evaporation of water, and attacks the boiler material by giving sodium ferroate (Na2FeO2), which decomposes and forms rust. Na 2 FeO FeO2 + 4 H 2 O  → 6N NaOH NaO aOH + F aOH Fee3O 4 + H 2 ↑ Sodium ferroatee

Rust R

This is an electrochemical phenomenon and can be explained on the basis that a concentration cell is formed due to concentration difference of sodium hydroxide in the boilers particularly at highly stressed parts like joints, rivets, etc. The dilute NaOH region in the boiler acts as a cathode and the concentrated NaOH region acts as an anode and undergoes corrosion. (+)

Iron at joint rivets, bends, etc.

Concentrated NaOH region

Dilute NaOH region

(−)

Iron at plane surfaces

Preventions (a) By using sodium phosphate as a softening agent instead of sodium carbonate. (b) By adding certain chemicals such as lignin and tannin to boiler water because they block the hair cracking inside the boiler. (c) By adding sodium sulphate to boiler water, which blocks the minute cracks thereby preventing the entry of sodium hydroxide solution.

1.15

SOFTENING OF WATER

In water, there is a formation of scale-like impurities in the boiler. This scale formation may be minimized by the following treatments: (i) Internal treatment (ii) External treatment

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(i) Internal treatment: In this method, chemicals are added to the water in the boiler, which is hard in nature. The added chemicals may function as precipitating agents or sequestering agents to form more soluble complex compounds with metal ions. In this method, hard deposit scale is changed into loose deposits, which are easily removed by blow-down operation. Internal treatment of boiler water depends on the nature of feed water and the type of the boiler. Some important internal treatment methods are: (a) Colloidal conditioning: In this method, scale formation can be reduced by introducing organic substances like kerosene, tannin, agar-agar, etc. They surround the minute particles of scale-forming salts, thereby yielding non-sticky and loose deposits, which can easily be removed by blow-down operation. (b) Carbonate conditioning: In low-pressure boilers, scale formation is prevented by adding sodium carbonate (Na2CO3) to boiler water to prevent the precipitation of scale-forming calcium sulphate (CaSO4). When calcium sulphate is converted into calcium carbonate by the addition of sodium carbonate, CaCO3 acts as a loose sludge, which can be removed by blow-down operation. CaSO 4 + Na 2 CO3 → CaCO CaCO3 + Na 2SO 4 loose sludge

Carbonate conditioning is not used in high-pressure boilers because excess of Na 2CO3 might be converted into NaOH due to hydrolysis, which causes caustic embrittlement. (c) Phosphate conditioning: Phosphate conditioning involves conversion of scale-forming calcium and magnesium salts into the most insoluble compound of calcium phosphate (Ca3(PO4)2) and magnesium phosphate (Mg3(PO4)2), which form easily removable nonadherent soft sludge, which can be removed by blow-down operation. 3CaCl 2 + 22Na Na 3 PO 4 → Ca 3 ( PO 4 )2 ↓ + 6 NaCl 3MgCl2 + 2Na 2 Na 3 PO 4 → Mg3 ( PO 4 )2 ↓ + 6 N NaCl aCl The three sodium orthophosphates may be used depending upon the alkalinity of the boilerfeed water. (1) In acidic medium, sodium dihydrogen phosphate (NaH2PO4) (2) In weakly alkaline medium, disodium hydrogen phosphate (Na2HPO4) (3) In alkaline medium, trisodium phosphate (Na3PO4) (d) Calgon conditioning: Addition of calgon or sodium hexameta phosphate (NaPO3)6 to boiler water converts calcium salts into soluble complex compound thereby preventing scale or sludge formation. Na 2 [ Na 4 (PO (PO3 )6 ] 22Na Na + + [Na [Na 4 ( P PO O3 )6 ]2 − Calgon

2CaSO O4 + [ N Naa 4 ( P PO O 3 )6 ]2 − → [Ca 2 (PO3 )6 ]2 − + 2Naa 2SO SO 4 soluble complex ion

(e) Sodium aluminate treatment: When we add sodium aluminate (NaAlO2) in boiler-feed water, it gets hydrolyzed and form sodium hydroxide (NaOH) and gelatinous precipitate of aluminium hydroxide Al(OH)3.

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1.40

Engineering Chemistry NaAlO O2 + H 2 O → A All(OH OH)3 + N NaOH gelatinous ppt

The sodium hydroxide reacts with magnesium salt and converts it into magnesium hydroxide Mg(OH)2 precipitates. MgCl2 + 2 NaOH → Mg(OH)2 + 2NaCl gelatinous ppt

The gelatinous precipitate of Al(OH)3 and Mg(OH)2 entraps colloidal and finely suspended impurities along with oil drops and silica. The loose slimy precipitate can be easily removed by blow-down operation. (f) Complexometric conditioning (EDTA conditioning): When EDTA is added to boiler-feed water having pH 8.5, then EDTA binds with the scale-forming cations to form stable and soluble complex. Hence, scale and sludge formation in boiler is prevented. (g) Electrical conditioning: In this method, sealed glass bulbs, containing mercury connected to a battery, are set rotating in the boiler. As water boils, mercury bulb emits electrical discharges, which prevent scale-forming particles to adhere together to form scale. (h) Radioactive conditioning: Radioactive salts containing tablets are placed inside the boilerfeed water at a few points. Energy radiated from radioactive substances prevents the scale and sludge formation. (ii) External treatment: Hard water causes a number of harmful effects when used for domestic, industrial, and boiler purposes. So we have to remove or reduce hardness-causing impurities present in water before using it for any purpose. The most common methods for softening of water are given below: (a) Lime soda process (b) Zeolite process/permutit process/base exchange process (c) Demineralization/ion-exchange process/de-ionization (a) Lime Soda Process It is a very important and popular process for softening of water. Principle: This method involves the treatment of water sample with calculated quantities of lime [Ca(OH)2] and soda (Na2CO3), which react with calcium and magnesium salts to form insoluble precipitates as calcium carbonate (CaCO3) and magnesium hydroxide (Mg(OH)2). To accelerate the precipitation of CaCO3 and Mg(OH)2, certain substances are added, known as “coagulants” or “flocculants.” Functions of lime: For removing temporary hardness, permanent magnesium hardness, free mineral acids, iron and aluminium salts, dissolved CO2 and H2S in water, lime acts as a good agent. (1) Removal of temporary hardness: Here lime converts bicarbonates into carbonates Ca( HCO HCO3 )2 + Ca Ca(OH OH)2  → 2C CaC aCO3 ↓ + 2H 2 O aCO Mg( HCO HCO3 )2 + 2Ca Ca(OH OH)2  → 2C CaC aCO3 ↓ + Mg aCO Mg(OH OH)2 + 2H 2 O

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(2) Removal of permanent magnesium hardness: Lime can remove the permanent magnesium hardness and converts the magnesium hydroxide MgCl2 + Ca(OH)2    → Mg(OH)2 ↓ + CaCl CaCl 2 MgSO4 + Ca(OH)2    → Mg(OH)2 ↓ + CaSO 4 In this case, permanent magnesium hardness is converted to permanent calcium hardness. (3) Removal of dissolved iron and aluminium salts: Lime can convert iron and aluminium salts to the corresponding hydrates. FeSO 4 + Ca(OH)2    → Fe(OH)2 ↓ + CaSO CaSO 4 Al 2 (SO 4 )3 + 3Ca(OH)2  → 2Al(OH)3 ↓ + 3CaSO 4 (4) Removal of dissolved CO2 and H2 S: Lime can remove the carbon dioxide as calcium carbonate and hydrogen sulphide as calcium sulphide. CO2 + Ca(OH)2    → CaCO CaCO3 ↓ + H 2 O H 2S + Ca(OH)2    → CaS CaS + 2H 2 O Functions of soda: When lime is used to remove the hardness or mineral acids, it has been found that permanent calcium hardness (CaCl2) and (CaSO4) is introduced in water. Soda is very effective to remove permanent calcium hardness as follows: CaCl 2 + Na 2 CO3 → CaCO CaCO3 ↓ + 2 NaCl CaSO 4 + Na 2 CO3 → CaCO CaCO3 ↓ + Na 2SO 4

Important Points about Calculation of Lime and Soda (i) There is no consideration of substances like NaCl, KCl, Na2SO4, SiO2, Fe2O3, etc., for calculating lime and soda requirement as they do not impart any hardness. (ii) Equivalent weight as NaAlO2 is equal to its molar mass. (iii) When the impurities are given as CaCO3 and/or MgCO3, they should be considered due to bicarbonates of calcium and/or magnesium, respectively. (iv) When the impurities are presented in the form of ions such as Ca2+ and/or Mg2+ ion, they are considered as a permanent hardness. (v) If there are OH− and CO32− ions present in the treated water, it indicates that excess of lime and soda, which are added for the treatment, and hence these excess amount should be added (in terms of CaCO3 equivalent) to the calculation. Requirement of lime and soda for the constituents responsible for hardness is given in Table 1.4.

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Engineering Chemistry Table 1.4 Calculation of lime and soda requirement Constituents

Chemical reaction

Requirement

Temporary Hardness Ca(( HCO HCO3 )2

Ca( HCO HCO3 )2 + Ca Ca(OH OH)2 → 2C CaCO CaC aCO3 + 2H 2O aCO

Mg(( HCO HCO3 )2

Mg(HCO3 )2 + 2C Ca(OH) a(OH)2 → 2CaCO3 + Mg(OH)2 + 2H 2O

Ca 2 + + Na 2CO3 → CaCO3 + 2 Na + Mg 2 + + Ca(OH)2 → Mg(OH)2 + Ca 2 +   Ca 2 + + Na 2CO3 → CaCO 3 + 2 Na + 

L+S

HCO O3− + Ca Ca(OH OH) 2 → C CaCO CaC aCO O3 + H 2O + C CO O32 −

L−S

2H + + Ca(OH)2 → Ca 2 + + 2H 2O   Ca 2 + + Na 2CO3 → CaCO3 + 2 Na + 

L+S

CO2 + Ca(OH)2 → CaCO3 + H 2O

2Al3+ + 3Ca 3Ca(OH)2 → 3Ca 2 + + 22Al(OH)3 2+ 3Ca + 3Na 3Na 2CO3 → 3CaCO3 + 66Na Na + 3+ 2+ [∵ 2Al = 3C Caa ]

L+S

Fe2 + + Ca(OH)2 → Fe(OH)2 + Ca 2 + Ca 2 + + Na 2CO3 → CaCO3 + 2 Na +

L+S

NaAlO O 2 + 2H 2O → A All(OH OH)3 + N NaOH [∵2 NaOH = Ca(OH)2 ]

−L

Permanent Hardness Ca 2+ Mg

Other Species HCO3− (e.g. g. NaHCO NaHCO3 )

H+ (Free Acid HCl, H 2SO 4 etc) CO2 Coagulants Al 2 (SO 4 )3

FeSO 4 NaAlO2

Formula for Lime and Soda Requirement 100 parts by mass of CaCO3 are equivalent to (i) 74 parts of Ca(OH)2 (ii) 106 parts of Na2CO3 Lime Requirement =

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74 Temp. Ca 2 + + 2 × Temp Mg2 + + per per( M Mgg2 + + Fe2 + + Al3+ ) 100  + H + + CO2 + HCO3− − N NaAlO aAlO2  all in terms of CaCO3 eq × Voll. oof water × purity factor (100/% purity)

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106  perm(Ca 2 + + Mg2 + + Al3+ + Fe2 + ) + H + − HCO3−  100  all in terms of CaaC CO 3 × V Vol.of water × purity factor (100/% purity)

Process: Lime soda process is of two types: (i) Cold lime soda process (at room temperature) (ii) Hot lime soda process (at 90°C–100°C temperature) (i) Cold lime soda process (a) Calculated quantity of lime and soda is mixed with hard water at room temperature. (b) At room temperature, the precipitates formed are fi nely divided, so they do not settle down easily. (c) So it is essential to add a small amount of coagulant, which hydrolyzes to form flocculent and gelatinous precipitate of aluminium hydroxide or ferric hydroxide, which entraps the fi ne precipitates (as shown in Figure 1.10). Hard water (containing Ca2+, Mg2+, or other heavy metals) + lime + soda (i) Addition of coagulants or flocculent (ii) Proper setting time ↓ ppts of CaCO3 + Mg(OH)2 settle out. FeSO 4 + Ca(OH) Ca(OH)2 → Fe(OH)2 + CaSO CaSO 4 Al 2 (SO (SO 4 )3 + 3Ca(HCO Ca(HCO3 )2 → 22Al(OH)3 ↓ + 33C CaSO O 4 + 6CO 6CO2 ↑ Aluminium sulphate

Gelantinous

(Coagulent)

ppts

Motor

Driving belt

Chemicals feed inlet (L + S + coagulants)

Raw water (inlet) Filtered softened water (outlet)

Stirrer paddles

Outer chamber Stirrer

Inner chamber

Sedimented sludge [CaCO3, Mg(OH)2] Sludge outlet

Figure 1.10 Cold lime soda softener

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Engineering Chemistry (d) Use of sodium aluminate (NaAlO2) as coagulant also helps in the removal of silica as well as oil, if present in water. NaAlO O2 + 2H 2H 2 O → NaOH + Al(OH)3 ↓

Sodium aluminate

Aluminium hydroxide

(e) The residual hardness after cold lime soda process is 50 to 60 ppm. (ii) Hot lime soda process (a) In this method, softening of water by lime and soda at temperatures close to the boiling temperature of water (100°C). (b) The chemical reactions proceed at a faster rate, because the viscosity of water is low at higher temperature and precipitates sludge settle down easily. (c) No coagulants are required for hot lime soda process. (d) The hot L-S plant consists of three parts (as shown below in Figure 1.11): (1) A ‘reaction tank’ in which feed water, chemicals, and steam are thoroughly mixed. (2) A ‘conical sedimentation vessel’ in which sludge settles down. (3) A ‘sand filter’ that ensures complete removal of sludge from the water. In the hot process, sodium carbonate (Na2CO3) is used for softening because it decomposes into sodium hydroxide under high pressure and temperature. Na 2 CO3 + H 2 O → 2 NaOH + CO2 Ca( HCO HCO3 )2 + 2 N NaOH NaO aOH → C aOH CaCO CaC aCO O3 ↓ + N Naa 2 CO CO 3 + 2 H 2 O CaCl 2 + Na 2 CO3 → CaCO3 ↓ + 2 NaCl Mg( HCO HCO3 )2 + 4 NaO NaOH aOH → M Mgg(OH OH)2 ↓ + 2 N Naa 2 C CO O3 + 2H 2 O MgCl2 + 2 NaOH → Mg( Mg(OH)2 + 2NaCl 2 NaCl (e) The residual hardness after hot lime soda process is 15–30 ppm. Raw water (Inlet)

Super heated steam (Inlet)

Chemical feed (L + S) Inlet

Reaction tank Conical sedimentation tank Precipitated sludge [CaCO3, Mg(OH)2]

Fine sand layer Coarse sand layer Gravel layer Filtered softened water (Outlet) Sludge outlet

Figure 1.11 Hot lime soda process

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Advantages of Lime Soda Process (i) It is very economical. (ii) Hot L-S process is much faster than cold lime soda process. (iii) During lime soda process, pH value of water is increased; hence corrosion of pipe is reduced. (iv) The alkaline nature of treated water controls the amount of pathogenic bacteria in water. (v) This process also helps to remove Fe and Mn to some extent. (vi) It removes not only hardness-causing salts but also other minerals. Disadvantages of Lime Soda Process (i) Soft water contains 15–30 ppm residual hardness. (ii) It requires careful operation and skilled supervision for efficient softening. (iii) Sludge disposal is different and poses a problem. Difference between cold and hot lime soda process is given in Table 1.5. Table 1.5 Difference between cold and hot lime soda process Cold lime soda process

Hot lime soda process

1. 2. 3. 4. 5. 6.

Very efficient process Reactions are fast and complete High softening capacity Filteration is fast Coagulants are not required Dissolved gases such as CO2 and H2S are removed to some extent Steam is essential Residual hardness of about 15–30 ppm

Less efficient process Reactions are comparatively slow Low softening capacity Filteration is slow Coagulants are essential Dissolved gases are not removed

7. Steam is not required 8. Residual hardness is around 60 ppm

(b)

Zeolite or permutit process Zeolite is known as permutit, i.e., boiling stone. Zeolite process is widely used to soften water. Zeolites are hydrated alumino silicate minerals. or Sodium aluminium orthosilicate, and it is represented as Na2O·Al2O3·xSiO2·yH2O (x = 2 – 10, y = 2 – 6) represented as Na2Z. Zeolites are of two types: (1) Natural zeolite: It is non-porous and derived from green sand. Example: Natrolite (Na2OAl2O3·3SiO2·2H2O) (2) Synthetic zeolite: It is porous and possesses a gel structure. It is prepared by heating china clay, feldspar, and soda ash together. They have higher exchange capacity as compared to natural zeolite. So it is more common in use. Softening Process In the zeolite process for softening hard water, the raw water is percolate through a bed of zeolite (Na2Z), which is packed in a vertical cylindrical tank as shown in the Figure 1.12. The zeolite bed retains the Ca2+ and Mg2+ ions from hard water by exchanging with Na+ ions thereby the out-flowing water contains sodium salts.

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Engineering Chemistry Feed water + Zeolite bed Regeneration with brine

Water softening Exhausted zeolite bed

Raw water (inlet)

Zeolite bed Gravels Injector

To sink Soft water (outlet) NaCl solution storage

Figure 1.12 Zeolite softener CaSO 4 + Na 2 Z → C CaZ ↓ + Na 2SO 4 MgSO4 + Na 2 Z → M MgZ ↓ + Na 2SO 4 CaCl 2 + Na 2 Z → C CaZ ↓ + 2 NaCl Ca( HCO HCO3 )2 + Na 2 Z → CaZ ↓ + 2 NaHCO3 Na2Z (Zeolite Bed) Mg(HCO3)2/

Ca(HCO3)2/ CaCl2/

Feed W r Wate

CaSO4

CaZ + 2NaHCO3/2NaCl/Na2SO4

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Feed Water

MgCl2/

MgSO4 MgZ + 2NaHCO3/2NaCl/Na2SO4

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This process removes both temporary and permanent hardness. After long use, the zeolite bed gets exhausted. It can be regenerated by using chemicals. Regeneration: When the zeolite bed is completely converted into calcium and magnesium zeolite, it no longer works as softener. It gets exhausted. At this stage, supply of feed water is stopped, and the exhausted zeolite is regenerated by treating with a concentrated (10%) brine (NaCl) solution. CaZ/MgZ Z + 2N NaCl NaC aCll → N Naa 2 Z + CaCl CaCl 2 /MgCl 2 Exhausted

Brine

Regenerated Zeollite ite

Zeolite

The regenerated zeolite bed thus obtained is used again for softening operation. Zeolite process reduces hardness to 0–15 ppm. Limitations (i) Hard water should be turbidity free otherwise impurities will clog the pores. (ii) Mineral acids must be removed if present because they destroy the zeolites. (iii) If Mn2+ or Fe2+ ions are present in feed water, it must be removed, otherwise they form MnZ or FeZ, which cannot be removed easily during regeneration. Advantages of Zeolite Process (i) Hardness-causing ions are completely removed with a very low residual hardness of about 10 ppm in the softened water. (ii) Zeolite process automatically adjusts for any variation in hardness of incoming water. (iii) It is a clean process because it does not produce any sludge. (iv) Zeolite equipment requires less area. (v) It requires less time for softening the water. (vi) It requires less skill for operation as well as maintenance. Disadvantages of Zeolite Process (i) (ii) (iii) (iv)

Zeolite process cannot tolerate acidity as the zeolite disintegrates. Turbid water cannot be treated satisfactorily. Treated water contains more sodium salts. This process replaces Ca2+ and Mg2+ ions by Na+ ions, and hence softened water contains more sodium and also more dissolved salts. (v) Anions such as HCO O3− , CO32 − remain in water as sodium salts, which contributing to the alkalinity causes corrosion and caustic embrittlement of the boiler. Difference between zeolite and lime soda process is given in Table 1.6. (c)

Demineralization process/Ion exchange process In this process the cations and anions present in water and which can produce hardness are removed by ion-exchange resins. Resins are long, cross-linked organic polymers with a porous structure. Ion-exchange resins are mainly (1) cation-exchange resins and (2)anionexchange resins.

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Engineering Chemistry Table 1.6 Differences between zeolite and lime soda process Zeolite process

Lime soda process

1. Softened water contains residual hardness about 10 ppm.

Softened water contains 50–60 ppm in cold lime soda process and 15–30 ppm in hot lime soda process. Treated water contains less amount of sodium salt. Capital cost is low. There is no such type of limitations.

2. Treated water contains large amount of sodium salt than in original raw water. 3. Cost of plant and material is higher. 4. This process is not useful for removal of acidic impurities, turbidity, etc. 5. The plant occupies less space. 6. Treated water contains more NaHCO3, which creates problem in boiler. 7. The raw water must be free from suspended impurities. Otherwise the pores of the zeolite get blocked. 8. It automatically sets itself to waters of different hardness. 9. This process removes only Ca2+ and Mg2+ ions in solution. It does not remove anions (HCO3− , Cl Cl −, SO24 − ) from solution. 10. It is not fit for industrial purposes especially in boiler.

The plant occupies more space. Treated water is free from NaHCO3, so it is used in boiler. This process has no such limitation. Frequent control and adjustment of reagent is needed. It removes cations and anions from the solution. It is fit for industrial purposes especially in boiler.

(1) Cation-exchange resins: These are styrene-divinylbenzene copolymers. These resins have acidic functional groups such as -COOH, -SO3H, etc., which are capable of exchanging the cation by their hydrogen ions. Hence, they are also called cation exchangers. They can be represented as R-H, where R is the insoluble polymeric heavy part. CH2

CH2

−

SO3H

−

SO3H CH

−

SO3H

CH2 CH2

CH2

H2C H 2C

CH2

−

SO3H

Sulphonation form of cation-exchange resin

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(2) Anion-exchange resins: These are copolymers of styrene and divinylbenzene containing basic functional groups such as amine, substituted amine, quaternary ammonium groups, etc. They can be represented as R R′′-OH. ′-OH. CH2

CH2

CH2NMe3+OH

CH2

−

CH2NMe3 OH

CH2N + Me3 OH

−

CH2

H2C H2C

CH2

−

CH2

CH2N + Me3 OH

−

Process: Both cation exchanger and anion exchanger are inter-connected with a pipe as shown in the Figure 1.13. The hard water is fi rst passed through cation-exchange resin chamber, which removes all the cations (e.g., Ca2++ and Mg2++) from it, and equal amount of H+ ions are released from its column to water. 2R − H + Ca 2 + /Mg 2 + → R 2 Ca /R 2 Mg + 22H + Cation-exchange reaction

Water + CO2 Impure Water

CO2

Cation exchanger

Anion exchanger

Gravel

Acid for regeneration

Steam Jacket

Alkali for regeneration

Demineralized Water Degasifier

Pump Washing

Washing

Figure 1.13 Demineralization by ion exchangers

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Engineering Chemistry After passing through cation-exchange chamber, the hard water is now pumped to ‘anion-exchange resin’ chamber where all anions like Cl−, SO2− 4 , etc., are removed, and equal amount of OH− ions are released from this resin bed to water.  R ′OH + Cl − → R ′Cl + OH − 2− − 2R ′OH + SO 4 → R 2′ SO SO 4 + 2OH  Anion exchange reaction 2− 2R ′OH + CO3 → R 2′ CO CO3 + 2OH −  H+ and OH– ions released from reactions in equivalent amount get combined to produce water molecules. H + + OH OH − → H 2 O Thus, the treated water is completely free from cations as well as anions, so it is known as demineralized or deionized water. Regeneration of Resins After some time of usage (depending on water) of cation/anion exchange resins will exhaust, and it is most important to regenerate. Regeneration of cation-exchange resins: The cation-exchange resins are regenerated by addition of dil. HCl or H2SO4: R2C Caa + 2H +  → 2R − H + Ca 2 + ( from acid acid )

( resin resin ) ( washing washing ) +

R2M Mgg + 2H  → 2R − H + Mg2 + ( from acid acid )

( resin resin ) ( washing washing )

Regenerationof anion-exchange resin: The anion-exchange resins are regenerated by addition of dil. NaOH: R 2′ SO SO 4 + 2OH −    → 2R ′ − OH + SO24 − ( from from bas basee)

( resin resin )

( washing washing )

−

R ′Cl + OH  → 2R R′′ − OH + Cl − ( from from bas basee)

( resin resin )

( washing washing )

After regeneration of both resins, columns are washed with deionized water, and the washed product is passed to sink. Advantages (i) It produces water of very low hardness ((≈ ≈22 ppm). ≈ (ii) Highly acidic or highly alkaline water also can be softened. Disadvantages (i) The equipment is very costly. (ii) Expensive chemicals are required for regeneration. (iii) Turbid water decreases the efficiency of resins.

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Mixed-bed deionizer: As shown in Figure 1.14, mixed-bed deionizer contains a single cylindrical vessel with a mixture of a strong cation exchanger and a strong anion exchanger, and is the most efficient process than separate column exchanger process. Hard water which pass through the mixed bed contacts number of times with both exchangers and purifies the water. Purified water is having less than 1 ppm hardness and also this is a most widely used convenient method.

Raw water Inlet

Exhausted mixed bed ionizer

Anion exchanger (Low density density)

NaOH soln (dilute) H2SO4 soln (dilute) Cation exchanger (High density)

Mixed ionizer bed

Demineralized Compressed Water water air for backwashing

Washing to sink

Figure 1.14 Mixed-bed deionizer Regeneration of resins: When the resins get exhausted, the mixed bed is backwashed with water. The lighter anion exchangers get displaced to form an upper layer above the heavier cation exchangers. Then the anion exchangers are regenerated by passing NaOH solution from the top and then rinsed with deionized water. The lower layer of cation exchangers is generated by passing H2SO4 solution and is finally rinsed with deionized water. The two beds are then mixed again by forcing compressed air through it. Now the regenerated bed is ready for use again.

1.16

DESALINATION

Desalination or desalting involves the removal of dissolved salts (e.g., NaCl) from water. The salinity of water is due to dissolved NaCl and to a smaller extent of other inorganic salts. Natural saline water such as sea water contains more than 35,000 ppm while brackish water contains dissolved salts in the range 1000–3500 ppm. Desalination of saline water may be achieved by any of the two approaches: (i) Separating water from the saline water As in (a) Distillation or evaporation (b) Freezing (c) Solvent extraction (d) Reverse osmosis (ii) Separating the salt from the saline water As in (a) Osmionic process (b) Electrodialysis

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Engineering Chemistry

Reverse osmosis and electrodialysis are more important in large-scale operations, and operation and principle involved in reverse osmosis are discussed. Reverse Osmosis (RO) (Hyper Filteration) This technique worksbased on the principle of osmosis. Reverse osmosis is a process by which a solvent such as water is purified of solutes by being forced through a semipermeable membrane through which the solvent, but not the solute may pass. It is exactly opposite of osmosis and hence it is known as reverse osmosis. Generally the tendency of a fluid, i.e., water, to pass through a semipermeable membrane into a solution where the solvent concentration is higher, thus equalizing the concentrations of materials on either side of the membrane is known as osmosis. But when pressure is applied on the concentrated side, the solvent will flow in the reverse direction. Reverse osmosis uses 100–150 micron thick membrane made from cellulose acetate or polymeric membranes having pores in the range of 0.0001–0.001 μm in diameter; it allows only water to pass through it and not to the salt. The water molecules diff use through the membrane while the contaminants get concentrated in the effluent stream and are discharged. Process: In this process, a high pressure ((≈15–40 kg cm–2) is applied to the sea water or brackish water, which is to be treated (as shown in Figure 1.15). The semi-permeable membrane allows only the solvent molecule (pure water) to pass through it. Thus dissolved ionic and non-ionic solvents are left behind, and water get purified from salt. Generally, we use membrane made up of polymethacrylate and polyamide polymers for this process. Advantages (i) It removes ionic as well as non-ionic salts present in saline water. (ii) It is economical, compact, and very simple. Pressure (≈15−40 kg cm−2)

Piston

Sea water

Semi-permeable membrane

Pure water

Pure water (Outlet)

Figure 1.15 Reverse osmosis unit

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1.53

It is effective in removing colloidal matters also. It is suitable for converting sea water into drinking water. It has long life and is easy to replace the membrane within two minutes. The water obtained from the process can be used in high-pressure boilers.

Electrodialysis Electrodialysis is another efficient technique used for the desalination of saline water and is a membrane process. Principle: Under the influence of an electric potential across a salt water solution, charged ions move towards respective electrodes through ions and selective membrane. The membranes are cation or anion selective, which basically means that either positive ions or negative ions will flow through cation-selective membrane consisting of sulphated polystyrene, which allows only cations to flow through and rejects anions. However, anion-selective membrane consists of polystyrene with quaternary ammonia, which allows only anions and rejects cations. Multiple membranes alternatively allow cation or anions to flow through. Hence, with this method we can get fresh water from saline water. Process: The process is carried out in a special type of the cell called electrodialysis cell (as shown in Figure 1.16). It consists of two electrodes and ion selective membranes which are permeable to either cation or anion. The anode is placed near anion-selective membrane while the cathode is placed near the cation-selective membrane. The anion selective membrane has positively charged functional groups such as R4N+ and therefore allows negatively charged ions only to pass through them. Similarly, Sea Water at pressure 5 − 6 Kg/m2

Cation selective membrane

Cathode (−)

C +

A −

C +

A −

C +

A −

−

Anion selective membrane Anode (+) + +)

Pure water outlet Concentrated sea water outlet

Figure 1.16 Electrodialysis of sea water

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Engineering Chemistry

a cation-selective membrane has negatively charged functional groups such as RCOO− and allows only positively charged ions to pass through it. Saline water under a pressure of around 5−6 kg/m2 is passed from the top of the cell and it passes between membrane pairs. When an emf is applied across two electrodes the cations (Na+) present in salt water move towards cathode through cation selective membrane and anions (Cl−) move towards the anode through anion selective membrane. As a result, the concentration of ions in alternate compartments 2, 4, 6 etc. decreases, while it increases in the alternate compartment 1, 3, 5 etc. Thus water in the even number compartments becomes pure and is collected from the bottom of the cell. Similarly, water in the odd number compartment becomes rich in the saline water i.e. it becomes concentrated saline water. It is collected from a separate outlet at the bottom of the cell. Advantages (i) It is economical. (ii) It is convenient and may be applied at room temperature. (iii) It is most compact in size and requires only electricity for operation.

1.17 1.17.1

REVIEW QUESTIONS Fill in the Blanks

1. Hard water prevents lathering of soap due to the presence of the dissolved salts of _______ and _______ . [ [Ans.: calcium, magnesium] 2. _______ water has high quantity of organic matter. [ [Ans.: Lake] 3. Hardness is expressed in terms of _______ equivalent. [ [Ans.: CaCO3] 4. 1 ppm = _______ mg/L = _______ °Fr = _______ °Cl. [ [Ans.: 1, 0.1, 0.07] 5. _______ indicator is used in EDTA titration. [ [Ans.: EBT] 6. Caustic alkalinity in water is due to _______ and _______ ions. [ [Ans.: OH−, CO32− ] 7. Solubility of _______ in water decreases with increase of temperature. [ [Ans.: CaSO4] 8. The phenomenon of carrying of water along with impurities by steam is called _______ . [ [Ans.: carry-over] 9. _______ is an ideal chemical for the removal of dissolved oxygen. [ [Ans.: Hydrazine]

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10. The presence of residual _______ in boiler water causes caustic embrittlement. [ [Ans.: caustic, NaOH] 11. By adding _______ to boiler water, caustic embrittlement can be prevented. [ [Ans.: sodium sulphate] 12. Sodium aluminated is added as a _______ in purification of water. [ [Ans.: coagulant] 13. During lime soda process, calcium and magnesium ions impurities precipitate into _______ and _______ . [ [Ans.: CaCO3, Mg(OH)2] 14. Name of natural zeolite is _______ . [ [Ans.: natrolite, Na2O . Al2O3 . 4SiO2 . 2H2O] 15. In zeolite process, calcium and magnesium ions are replaced by _______ ions. [ [Ans.: sodium] 16. Exhausted cation-exchange column is regenerated by passing a solution of _______ . [ [Ans.: dilute HCl or dilute H2SO4] 17. Cation exchange resin have acidic functional groups like –SO3H, –COOH, –OH capable of exchanging cation by their _______ ions. [ [Ans.: hydrogen, H+] 18. Anion-exchange resins are having _______ ions, which are capable of exchanging anions in water. [ [Ans.: hydroxide, OH–] 19. _______ process produces least residual hardness in water. [ [Ans.: Ion-exchange] 20. _______ and _______ are used for colloidal conditioning. [ [Ans.: Tannin, agar-agar] 21. Water having no ions is called as _______ water. [ [Ans.: demineralized or deionized] 22. Calgon having chemical name is _______ . [ [Ans.: sodium hexameta phosphate, Na2[Na4(PO3)6]] 23. For domestic use, pH of water should be in the range of _______ . [ [Ans.: 7–8] 24. _______ flocculent is precipitated out when Al2(SO4)3 alum is added as a coagulant in water. [ [Ans.: Al(OH)3] 25. Chemical formula of alum is _______ . [ [Ans.: K 2SO4 . Al2(SO4)3 . 24H2O]

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26. _______ is the best way for disinfection of water in comparison with chlorine or bleaching powder. [ [Ans.: chloramines, ClNH2] 27. In electrodialysis process, _______ can be separated from brackish water. [ [Ans.: NaCl] 28. Electrodialysis consists of _______ selective membranes. [ [Ans.: ions] 29. Cation selective membrane is allowed to pass _______ and move towards cathode, and anion selective membrane is allowed to pass _______ and move towards anode. [ [Ans.: cations, anions] 30. Flow of solvent from a region of low concentration to high concentration when two solutions are separated by semi permeable membrane is called _______ . [ [Ans.: osmosis]

1.17.2

Multiple-choice Questions

1. The purest form of natural water is (a) River water (c) Underground water [ [Ans.: d]

(b) Sea water (d) Rain water

2. The alkaline hardness of water is due to the presence of the following salts of calcium and magnesium in water. (a) HCO3− only (b) HCO3− and CO32− only 2− (c) SO 4 only (d) HCO3− , CO32− , and OH− only [ [Ans.: d] 3. A sample of water contains 120 mg of Mg2+ per liter. The hardness of the sample of water in terms of CaCO3 equivalent is (a) 120 mg/L (b) 500 mg/L (c) 250 mg/L (d) 1000 mg/L [ [Ans.: b] 4. The total hardness of a sample of water is 1.88°Cl eq. CaCO3. Its hardness in ppm would be (a) 26.88 (b) 18.8 (c) 0.188 (d) 34.65 [ [Ans.: a] 5. The colour obtained by adding EBT indicator to a sample of water containing Ca2+ and Mg2+ at pH = 9–10 is (a) Blue (b) Wine red (c) Pink (d) No colour [ [Ans.: b]

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6. On boiling and filtering hard water, water sample contains (a) Temporary hardness (b) Permanent hardness (c) Both (d) None of the above [ [Ans.: b] 7. Blow-down operation causes the removal of (a) Scales (b) Sludges (c) Acidity (d) Basicity [ [Ans.: b] 8. Scale formation in boiler-feed water is due to (a) Metallic deposition (b) Corrosion in boilers (c) Deposition of hard water (d) All the above [ [Ans.: c] 9. Scale formation is mainly due to which of the following salt present in boiler-feed water? (a) CaSO4 (b) MgCO3 (c) Na2SO4 (d) KCl [ [Ans.: a] 10. Solubility of CaSO4 salt present in water (a) Increases with increase in temperature (b) Decreases with increase in temperature (c) Remain unchanged with increase in temperature (d) Not having any defi nite change with increase in temperature. [ [Ans.: b] 11. EDTA method is used for determining (a) Temporary hardness (c) Temporary and permanent hardness

(b) Permanent hardness (d) Alkalinity

[ [Ans.: c] 12. When phenolphthalein alkalinity, P = M then alkalinity is due to (a) OH− (b) CO32− (c) HCO3− [ [Ans.: b]

(d) CO32− and HCO3−

13. Permanent hardness of water cannot be removed by (a) Lime soda process (b) Permutit process (c) Boiling (d) Demineralization process [ [Ans.: c] 14. Permutit is chemically (a) Sodium silicate (c) Aluminium silicate [ [Ans.: b]

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(b) Hydrated sodium alumino silicate (d) All the above

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15. Hard water is not suitable for use in boilers because (a) It has a higher boiling point (b) It leads to scale formation in the boiler (c) It consumes more fuel in steam generation (d) The quality of steam generated is not good [ [Ans.: b] 16. Which of the following substances is capable of removing dissolved oxygen from water? (a) Cl2 (b) N2H4 (c) Na2SO4 (d) CaOCl2 [ [Ans.: b] 17. Sterilization of water can be done by using (a) H2O2 (b) O2 (c) Cl2 (d) NaOH [ [Ans.: c] 18. Coagulants help in settling of (a) Suspended impurities only (b) Finely suspended impurities only (c) Colloidal particles only (d) Both the suspended and colloidal particles [ [Ans.: d] 19. 1 ppm of K+ present in a sample of demineralized water is equal to (a) 4.3478 × 10 –8 mol L –1 (b) 2.564 × 10 –10 mol L –1 –5 –1 (c) 2.564 × 10 mol L (d) None of the above [ [Ans.: c] 20. Calgon is a name given to (a) Sodium silicate (b) Sodium hexameta phosphate (c) Sodium meta phosphate (d) Calcium phosphate [ [Ans.: b] 21. Permutit exchanges Ca2+ and Mg2+ ions present in hard water with (a) Zeolite ions (b) H+ ions + (c) Na ions (d) None of these [ [Ans.: c] 22. The exhausted zeolite can be regenerated by treating it with (a) 10% NaCl solution (b) 50% NaCl solution (c) 10% HCl solution (d) 50% HCl solution [ [Ans.: a]

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Water Technology 23. Brackish water contains dissolved (a) CaSO4 (c) NaCl [ [Ans.: c]

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(b) MgCl2 (d) Na2SO4

24. Which of the following samples of water cannot be softened by zeolite process? (a) Water containing temporary hardness (b) Water containing permanent hardness (c) Water containing excess of alkalinity (d) Water containing excess of dissolved salts [ [Ans.: c] 25. The cation-exchange resins possesses (a) Acidic groups (c) Amphoteric groups [ [Ans.: a]

(b) Basic groups (d) None of these

26. Priming and foaming process in boiler-feed water is due to (a) The formation of air bubbles and production of wet steam (b) The formation of scales (c) The formation of sludges (d) None of these [ [Ans.: a] 27. The cation and anion resins are made up of the basic polymer unit of (a) Polyvinyl chloride (b) Poly acrylate (c) Poly styrene (d) Polybutadiene [ [Ans.: b] 28. By ion-exchange process the hardness of water can be reduced up to (a) 0 ppm (c) 10 ppm

(b) 5 ppm (d) 15 ppm

[ [Ans.: a] 29. Boiler corrosion caused by using highly alkaline water in boiler is called (a) Corrosion (b) Boiler corrosion (c) Caustic embrittlement (d) Erosion [ [Ans.: c] 30. Desalination is the process of removing (a) Common salt from sea water and making it potable (b) Hard salts from sea water (c) NaOH from hard water (d) None of the above [ [Ans.: a]

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31. An exhausted anion-exchange resin can be regenerated by treating it with (a) Conc. HCl solution (b) Conc. NaOH solution (c) Dilute brine solution (c) Conc. brine solution [ [Ans.: b] 32. Tannins and agar-agar are used for (a) Phosphate conditioning (c) Radioactive conditioning [ [Ans.: b]

(b) Colloidal conditioning (d) Calgon conditioning

33. Alum is commonly used in the treatment of municipal water for (a) Filteration (b) Sedimentation (c) Coagulation (d) Flocculant [ [Ans.: d] 34. The chemical formula of alum is (a) K 2SO4 . Al2 (SO4)3 . 20H2O (c) K 2SO4 . Al2 (SO4)3 . 24H2O [ [Ans.: c]

(b) KNO3 . Al2 (SO4)3 . 24H2O (d) K 2SO4 . Al2 (SO4)3 . 21H2O

35. Liquid chlorine is most effective (a) Disinfectant (c) Flocculant [ [Ans.: a]

(b) Coagulant (d) Sterilizing agent

36. The soft, loose, and slimy precipitate formed within the boiler is called (a) Scale (b) Sludge (c) Flocculant (d) Coagulant [ [Ans.: b] 37. In reverse osmosis (RO) the flow of solvent is due to (a) Potential gradient (b) Vapour pressure gradient (c) Concentration gradient (d) None of the above [ [Ans.: c] 38. In RO process, the membrane used is (a) Polysulfone (c) Poly amide [ [Ans.: d]

(b) Polysulfone amide (d) All above

39. Chemical formula of bleaching powder is (a) Cl2 (c) NH2Cl [ [Ans.: d]

(b) HOCl (d) CaOCl2

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40. The acid responsible for disinfection of germs and bacteria is (a) HCl (b) HNO3 (c) HOCl (d) H2CO3 [ [Ans.: c]

1.17.3

Short Answer Questions

1. Name the main sources of water. Ans.: Sea water, rain water, ground water, and surface water. 2. What is the cause for alkalinity of natural water? Ans.: Due to the presence of dissolved bicarbonates of Ca and Mg in water. 3. Defi ne hardness of water. Ans.: Hardness is the characteristic property, which produces white scum on treating with soap solution. 4. Why does not hard water give lather with soap? Ans.: Because hard water produces insoluble white precipitate on treating with soap. 2C17H35COONa + CaCl2 → (C17H35COO)2 Ca↓ + 2NaCl Soap Hardness White Scum 5. How is hardness of water expressed? Ans.: The concentration of hardness-producing salts is expressed in terms of calcium carbonate (CaCO3) equivalent. 6. How hardness is determined in terms of CaCO3 equivalent. Ans.: Hardness (CaCO3 equivalent) =

W × 50 E

W = Weight of hardness-producing substance in ppm E = Equivalent weight of hardness-producing substance 7. Defi ne ppm, mg/L, Clarke’s degree, and French degree. Ans.: ppm: 1 part of CaCO3 equivalent hardness present in 106 parts of water. mg/L: Number of mg of CaCO3 equivalent hardness present in 1L of water. Clarke’s degree: Number of parts of CaCO3 equivalent hardness present in 70,000 parts of water. French degree: Number of parts of CaCO3 equivalent hardness present in 105 parts of water. 8. What is the relationship between ppm, mg/L, °Cl, and °Fr. Ans.: 1ppm = 1mg/L = 0.07°Cl = 0.1°Fr 9. Explain why water containing Ca2+ (aq) and HCO3− (aq) ions is said to be hard. Ans.: The Ca2+ ions give precipitates with soaps. On heating HCO3− ions, they are converted to CO32− ions, which precipitate in kettles/boilers with Ca2+ ions. 10. Why do we express hardness of water in terms of CaCO3 equivalent? Ans.: Because addition and subtraction of concentration of hardness-causing constituents are easy. Its molecular mass is 100.

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11. What are the salts responsible for the temporary and permanent hardness of water? Ans.: Temporary hardness: Mg(HCO3)2 and Ca(HCO3)2 Permanent hardness: CaCl2, MgCl2, CaSO4, MgSO4, FeSO4, etc. 12. Name the gases that dissolve in water and cause corrosion. Ans.: Oxygen, carbon dioxide, and sulphur dioxide. 13. What happens when hard water is boiled? Ans.: On boiling, temporary hardness is removed by precipitating as

∆ → CaCO3↓ + H2O + CO2↑ Mg(HCO3)2 ∆ → Mg(OH)2↓ + H2O + CO2↑ Ca(HCO3)2

14. Name any three substances that are used for sterilization of water. Ans.: (i) Liquid chlorine (ii) Bleaching powder (iii) Chloramine 15. Why is chlorination is better than chlorine or bleaching powder for sterilization of water. Ans.: Because chloramine (i) is quite stable (ii) does not impart bad taste to treated water (iii) imparts good taste to treated water 16. What is break-point chlorination? Ans.: It involves addition of sufficient amount of chlorine to water in order to (i) oxidize organic matter (ii) reduce substance and (iii) free ammonia and leaves behind mainly free chlorine for disinfecting disease-producing bacteria. 17. What are the advantages of break-point chlorination? Ans.: (i) It oxidizes organic matter, NH3, and reducing substances completely. (ii) It removes colour in water. (iii) It destroys all the disease-producing bacteria completely. (iv) It removes odour from water. (v) It prevents any growth of weeds in water. 18. Mention the impurities present in natural water. Ans.: (i) Suspended impurities (ii) Colloidal impurities (iii) Dissolved impurities 19. What is standard hard water. Ans.: Usually it is a solution containing 1 g of CaCO3 equivalent hardness in 1 liter, i.e., 1000 ppm of hardness water. 20. What is sedimentation with coagulation? Ans.: The process of removing of fi nely suspended impurities as well as colloidal impurities by adding requisite amount of coagulant to water before sedimentation.

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21. What is colloidal conditioning? Ans.: Scale formation can be avoided in low-pressure boilers by adding substances like kerosene, tannin, agar-agar, etc., which get adsorbed over the scale-forming precipitates, thereby yielding non-sticky and loose deposits, which can be removed by blow-down operation. 22. What is the indicator used in EDTA method? What is the end-point? Ans.: Indicator: EBT End point: Wine red to pure blue 23. Why is NH 4 OH-NH4Cl buffer solution added during the determination of hardness of water by EDTA method? Ans.: The indicator used in this titration (EBT) shows colour change at a pH value of about 10. So alkaline buffer (NH 4 OH-NH4Cl) is used. 24. Soft water is not demineralized,whereas demineralized water is soft. Why? Ans.: Soft water may contain Na+, Cl–, and SO2− 4 ions, so it is not demineralized, whereas demineralized water does not contain any cation and anion. 25. Why is water softened by zeolite process fit for use in boilers? Ans.: Because zeolite-softened water contains large quantities of sodium salts like NaCl, Na2SO4, etc., which avoids caustic embrittlement. 26. CO2 should not be present in boiler-feed water. Why? Ans.: Because CO2 forms carbonic acid (H2CO3) on reacting with water. So boiler’s wall material can be attacked slowly by carbonic acid and becomes weaker and weaker progressively. 27. What is meant by softening of water? Ans.: Softening of water means removing hardness-producing salts from water. 28. Why is water softened before using in boilers? Ans.: Water should be softened before using in boilers otherwise it may cause various boiler problems like (i) scale and sludge formation (ii) priming and foaming (iii) boiler corrosion. 29. What is meant by disinfection of water by UV method. Ans.: When water is irradiated by UV radiations, microorganisms and bacteria are killed. This so-called disinfection of water by UV radiation. 30. What is zeolite? Ans.: It is hydrated sodium alumino-silicate having formula Na2O . Al2O3 . xSiO2 . yH2O, where x = 2–10 y = 2–6 It is represented as Na2Z, and Na+ ions are capable of exchanging by M2+ (Ca2+ or Mg2++) present in water sample. 31. If silica is present in water, what harmful effects it can cause to boilers? Ans.: If silica is present in water, it causes formation of very fi rmly sticking deposits of calcium silicate (CaSiO3) and magnesium silicate (MgSiO3) scales in the boilers, which are very difficult to remove.

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32. Why is the presence of NaAlO2 in water equivalent to the presence of equivalent to Ca(OH)2? Ans.: NaAlO2 + H2O→Al(OH) → →Al(OH) 3 + NaOH 1 eq. 1 eq. of Ca(OH)2 33. Alkalinity of water cannot be due to presence of (a) OH −, CO 32− and HCO3 − or (b) OH −, HCO3− in water. Give reasons. Ans.: Because, OH− and HCO3− react to form CO32− OH − + HCO3− → CO32− + H2O. 34. Why does magnesium bicarbonate require double amount of lime for softening? Ans.: Mg(HCO3)2 + 2Ca(OH)2 → 2CaCO3↓ + Mg(OH)2↓ + 2H2O Thus from the above equation, mole of Mg(HCO3)2 ≡ 2 mol of Ca(OH)2 35. Are coagulants also used in hot lime soda process? Give reasons. Ans.: No, because reaction proceeds faster in hot lime soda process, and the precipitate and sludge formed settle down rapidly. Thus, no coagulants are required in hot lime soda process. 36. Water should not be soft for drinking purposes. Why? Ans.: Water should not be soft for drinking purposes because soft water is plumbosolvent, i.e., it attacks lead used in plumbing. 37. What is the main advantage of reverse osmosis over ion-exchange process? Ans.: Reverse osmosis removes all ionic, non-ionic, colloidal, and high molecular weight organic matter. 38. Why does the water softened by lime soda process cause boiler troubles? Ans.: The treated water still contains some residual hardness. 39. Why can caustic embrittlement be controlled by adding Na2SO4 to boiler-feed water? Ans.: When Na2SO4 is added to boiler-feed water, it blocks hair cracks, thereby preventing infilteration of caustic soda solution in these areas. So by this way, caustic embrittlement is prevented by using Na2SO4 in boiler-feed water. 40. Why is calgon conditioning better than phosphate conditioning? Ans.: In calgon conditioning, the added calgon forms soluble complex compound with CaSO4, thereby it prevents the scale and sludge formation in water. + 2– Na2[Na4(PO3) 6] 2Na + [Na4(PO3)6]

calgon

2CaSO4 + [Na4(PO3)6]2– → [Ca2(PO3)6]2– + 2Na2SO4 soluble complex

This soluble complex does not cause any problem in the boilers. On the other hand, in phosphate conditioning, sodium phosphate is added to boiler water so that precipitate of calcium phosphate is formed. Although this precipitate is nonadherent and soft, it has to be removed by frequent blow-down operation. 2Na3PO4 + 3CaSO4 → Ca3(PO4)2↓ + 3Na2SO4 Hence, calgon conditioning is better than phosphate conditioning.

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Water Technology

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Descriptive Questions

Q.1 Complete the following equations: a. 1 ppm = ______mg/L = ______°French = ______°Clark ∆ → b. Ca(HCO3)2  c. Mg(HCO3)2 + 2Ca(OH)2 → d. Na2[Na4(PO3)6] + 2CaSO4 → Q.2 Explain the following: a. Scale and sludge formation and their disadvantages b. Caustic embrittlement c. Boiler corrosion Q.3 What is hardness of water? How is it determined by EDTA method? Q.4 Describe the continuous lime soda process of softening hard water. Compare continuous cold lime soda process with hot lime soda process. Q.5 How is true exhausted zeolite bed regenerated? Give the merits and demerits of zeolite process. Q.6 What are the requirements of water for domestic use? Q.7 A water sample contains Ca(HCO3)2 = 32.4 mg/L, Mg(HCO3)2 = 29.2 mg/L, and CaSO4 = 13.5mg/L. Calculate the temporary and permanent hardness. [ [Ans.: 40 ppm,10 ppm] Q.8 Calculate the hardness of water containing the following salts: CaSO4 = 28 mg/LMg(HCO3)2 = 22 mg/L MgCl2 = 30 mg/LCaCl2 = 85 mg/L [ [Ans.: Temporary hardness = 15.07 ppm Permanent hardness = 128.7 ppm] Q.9 1 g of CaCO3 was dissolved in dil. HCl, and the solution was diluted to 1 liter. 50 ml of this solution required 45 ml of EDTA solution. 50 ml of hard water required 18 ml of EDTA solution during titration in ammonia buffer using EBT indicator. On the other hand, 50 ml of boiled water sample required 9 ml of EDTA solution under the same condition. Calculate each type of hardness in ppm. [ [Ans.: Total hardness = 400 ppm Permanent hardness = 200 ppm Temporary hardness = 200 ppm] Q.10 0.28 g of CaCO3 was dissolved in HCl and the solution made up to 1 liter with distilled water. 100 ml of the above solution required 28 ml of EDTA solution on titration. 100 ml of a hard water sample required33mlof same required solution on titration. After 100 ml of this water, cooling and filtering and then titrated 10 ml of EDTA solution. Calculate the temporary and permanent hardness. [ [Ans.: 230 mg/L, 100 mg/L] Q.11 Explain the ion-exchange method of purifying the water. Discuss their use and regeneration, giving the reaction involved.

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Q.12 Write a short note on break-point chlorination. Q.13 Pure soft water is not fit for drinking purpose. Why? Q.14 Write the principle of lime-soda process? Why should we use coagulants along with lime and soda? Why is water softened by zeolite process that is unfit for use in boilers? Q.15 Explain reverse osmosis process for desalination of sea water. Q.16 A water sample contains the following impurities: Ca2+ = 20 ppm, Mg2+ = 18 ppm, HCO3− =183 ppm, and SO2− 4 = 24 ppm. Calculate the amount of lime and soda needed for softening. [ [Ans.: Lime = 185 mg/L Soda = Zero mg/L] Q.17 Water sample on analysis gave the following results: Mg(HCO3)2 = 70 mg/L, CaCl2 = 220 mg/L, MgSO4 = 120 mg/L Ca(NO3)2 = 164 mg/L. Calculate the quantity of lime (80% pure) and soda (90% pure) needed for softening the 10,000 liters of water. [ [Ans.: Lime = 1.81 kg Soda = 4.68 kg] Q.18 A water sample contains the following constituents in ppm: Mg(HCO3)2 = 73, MgCl2 = 95, MgSO4 = 12, CaSO4 = 68, Ca(HCO3)2 = 81, and NaCl = 4.8. Calculate the cost of chemicals required for softening 20,000 liters of water, if purity factor for lime is 95% and soda is 90%. The costs per 100 kg each of lime and soda are Rs. 75 and Rs. 2480, respectively. [ [Ans.: Lime cost = Rs. 3.03; Soda cost = Rs. 93.44] Q.19 What do you mean by pre-chlorination, post-chlorination, and superchlorination. Write the significance of break-point chlorination. Q.20 What do you mean by screening, sedimentation, and coagulant sedimentations? How are colloidal impurities removed from water? Q.21 What are the factors that cause alkalinity in water? How is alkalinity of water determined by titrimetric method ? Q.22 Write a short note on the followings: a. Phosphate conditioning c. Colloidal conditioning e. EDTA conditioning

1.17.5

b. Calgon conditioning d. Carbonate conditioning

Problems for Practice

1. 200 ml of water sample require 25 ml of N/50 H2SO4 for neutralization to phenolphthalein N end point. After that methyl orange was added to this, and further acid required was 35 ml 50 H SO . Calculate the type and amount of alkalinity of water as CaCO in ppm. 2

[ [Ans.: CO32− = 250 ppm HCO3− = 50 ppm]

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2. Calculate the amount of lime (88.3% pure) and soda (99% pure) required to soften 24,000 liters of water per day, which contains the following: CaCO3 = 1.85 ppm CaSO4 = 0.34 ppm MgCO3 = 0.42 ppm MgCl2 = 0.76 ppm MgSO4 = 0.90 ppm NaCl = 2.34 ppm [ [Ans.: Lime = 88.49 kg Soda = 46.25 kg] 3. Calculate the amount of lime and soda needed for softening 106 liters of water sample, which contains Mg2+ = 36 ppm, Ca2+ = 20 ppm, and HCO3− = 183 ppm. [ [Ans.: Lime = 222 kg Soda = 53 kg] 4. A water sample contains the following: Ca2+ = 120 ppm, Mg2+ = 120 ppm, CO2 = 132 ppm, HCO3− = 122 ppm, and K+ = 40 ppm. Calculate the amount of lime 80% pure and soda 90% pure for softening 106 liters of water sample. [ [Ans.: Lime = 832.5 kg Soda = 824.4 kg]

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POLYMERS “Important things for easy life—through polymers”.

2.1 INTRODUCTION In modern world, polymers are an integral part of an individual’s life. They have the most diverse structure and applications ranging from domestic articles to highly sophisticated instruments. Polymers are used in almost all fields such as medicine, industry, agriculture, construction and so on. In recent days, these materials are used to prepare nanomaterials. The human body is built up and functions with polymers such as DNA, RNA, hormones, enzymes, proteins, lipids, phosphonitrilic acids and so on. Most of the food materials that we eat are polymers such as carbohydrates, starch and so on. In view of their importance, a proper understanding of polymeric materials is very essential. The word polymer is derived from Greek word poly, which means many and meros which means units (or) parts. Polymers are macromolecules of high molecular masses built up by the linking together of a large number of small, repeated units by a covalent bond. The repeating unit present in the formation of a polymer is known as polymerisation. For example, Polymer

Structure

1. Polyethylene 2. Polyvinyl chloride

Monomer

[ CH2

CH2 ]n

H2C

CH2 (Ethylene)

[ CH2

CH ]n

H2C

CH (Vinyl chloride) Cl

2.2 DEGREE OF POLYMERISATION The size of the polymer molecule is decided by the number of repeating units present in it. The number of repeating units (n) in a chain formed in a polymer is known as the “degree of polymerisation”. H

Ethene (monomer)

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H C ]n H

Polythene (polymer)

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2.2 Engineering Chemistry In the figure, n is the degree of polymerisation. It is different from polymer and can be 104 or more. Molecular weight of polymer = Molecular weight of repeating unit × degree of polymerisation

2.3 CLASSIFICATION OF POLYMERS Polymeric materials can be classified into several ways:

2.3.1 Classification Based on Source (i) Natural polymers: These polymers are isolated from natural material such as animals and plants. For example, cotton, silk, wool, nucleic acids, proteins, starch, cellulose, natural rubber, etc. CH3 nCH2

CH3 CH

Polymerization

CH2

[ CH2

2-methyl 1-3 butadiene (isoprene)

CH2 ]n

Poly-isoprene (rubber)

(ii) Synthetic polymers: Polymers synthesised from low molecular weight compounds are called synthetic polymers. For example, polyethylene, polyvinyl chloride, nylon, terylene, etc. nH2C

Polymerization

[ H2C

CH ]n Cl

Cl Vinyl chloride

Polyvinyl chloride

2.3.2 Classification Based on Composition (i) Univalent or homopolymers: These are formed with the same monomer units. For example, polyethylene, polystyrene, polyvinyl chloride, etc. (ii) Copolymers: These are formed with two or more different monomer units. For example, Copolymer

Monomer molecules

1. Styrene butadiene rubber 2. Nitrile rubber

Styrene + Butadiene Acrylonitrile + Butadiene

Copolymers are classified into four categories depending upon the nature of the distribution of different monomers in the polymer chain. (a) Random copolymer: These are formed by the random arrangement of monomer units in the chain. A B A A A B B A B B (b) Alternating copolymers: Monomer units in a copolymer molecule are arranged in an alternate manner. A

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Polymers

2.3

(c) Block copolymer: A block copolymer consists of one in which blocks of repeating units of one monomer alternate with blocks another monomer. A

(d) Graft copolymer: This copolymer consists of a linear polymer chain of one monomer to which side chain of different monomer has been grafted.

B A

B B B

B B B B

2.3.3 Classification Based on Chemical Composition (i) Organic polymers: A polymer whose backbone chain is essentially made of carbon atoms is termed as organic polymer. For example, polyethylene, polystyrene, polyvinyl chloride, etc. (ii) Inorganic polymers: Polymers that are formed by non-carbon-carbon bonds are inorganic polymers. For example, polysilanes, polygermanes, etc.

2.3.4 Classification Based on Structure (i) Linear polymers: Here, monomeric units are joined in the form of long, straight chains. For example, nylons, polyester, polyvinyl chloride, high-density polythene, etc. (ii) Branched chain polymers: These are mainly linear in nature but also possess some branches along the chain. For example, glycogen, amylopectin, low-density polythene, etc. (iii) Cross-linked polymer: These polymers adjacent chains are joined one to another at various positions by covalent bonds. For example, elastomers like rubber. (iv) Network polymers: Monomer units that have trifunctional groups form three dimensional networks. The cross-linked polymer is also called network polymer. For example, Bakelite, urea formaldehyde resin, silicones, etc.

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2.4 Engineering Chemistry

2.3.5 Classification Based on Mode of Polymerisation (i) Addition polymers: Addition polymerisation takes place in compounds containing reactive double bonds. Chain polymerisation is characterised by a self-addition of the monomer molecules to each other very rapidly through a chain reaction. For example, polyethylene, polypropylene, polyvinyl chloride, etc. (ii) Condensation polymers: This type of polymerisation was brought about by monomers containing two or more reactive functional groups condensing with each other to form a large condensed polymer and also loss of small molecules such as H2O, NH3, HCL and so on. For example, polyester, nylon, polyamide, etc.

2.3.6 Classification Based on the Molecular Forces (i) Thermoplastic polymers: These are linear, long chain polymers, which can be softened on heating and hardened on cooling reversibly. Thus, they can be processed several times. For example, polythene, polypropylene, polyvinyl chloride, etc. (ii) Thermosetting polymers: These polymers get hardened during moulding, and once they have solidified, they cannot be softened. During moulding, such polymers acquire three-dimensional cross-linked structure with predominantly strong covalent bonds. For example, polyester, Bakelite, urea formaldehyde, etc. (iii) Elastomer: These are rubber-like elastic polymers, which can be stretched to at least thrice its length, but return to their original shape and dimension as soon as the stretching force is released. (iv) Fibres: Fibres are polymers whose chains are held by strong intermolecular forces like hydrogen bonding. They are crystalline in nature and of high tensile strength, due to strong intermolecular forces. For example, nylon, polyester, etc.

2.3.7 Classification Based on Tacticity The stereo chemical arrangement of the monomer units in the main chain of a polymer is known as tacticity. The orientation of monomeric units in a polymer can take an orderly or disorderly twist with respect to main chain polymers that are mainly classified into isotactic, syndiotactic and atactic polymers. (i) Isotactic polymers: In this polymer, all the side groups lie on the same side of the chain (cis arrangement), e.g., natural rubber.

R H C

C H

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R H C

H H

C H

R C

C H

C C

H H

R C

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Polymers

2.5

(ii) Syndiotactic polymer: In this polymer, side groups are arranged alternatively (trans arrangement), e.g., Gutta-percha.

H H C

C H

H C

H H

C H

H H

C H

C H H

C H

(iii) Atactic polymer: In this polymer, the monomers randomly arranged to the main chain, e.g., polypropylene.

R H

H H

C H

H H

C H

H C

C H H

C H

2.4 TYPES OF POLYMERIZATION Polymerisation is mainly of two types: (i) Condensation polymerisation (or) step polymerisation (ii) Vinyl or addition polymerisation (or) chain polymerisation

2.4.1 Condensation Polymerisation or Step Polymerisation Condensation is brought about by monomers containing two or more reactive functional groups condensing with each other to form large condensed polymer and also loss of small molecules such as H2O, NH3, HCl and so on. (i) Polyester: Condensation between carboxylic acid and diol gives polyester. 2nH

R OH + nHOOC

Diol

COOH

Dicarboxylic acid

O O

O R

+ 2(n − 1)H2O

Polyester

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2.6 Engineering Chemistry (ii) Polyethylene terephthalate (PET): CH2

n HO

CH2

OH + nHOOC

Ethylene glycol

H [O

CH2

COOH

Teryphthalic acid

CO ]n OH + 2(n − 1)H2O

OOC PET

(iii) Polyamide: Condensation reaction between diamine and dicarboxylic acid gives polyamide. H nH2N

NH2 + nHOOC

Diamine

COOH

[

Dicarboxylic acid

O R

C ]n + 2(n − 1)H2O

Polyamide

(iv) Nylon 66: Condensation reaction between hexamethylenediamine and adipic acid gives nylon 66. nH2N

NH2 + nHOOC

(CH2)6

Hexamethylenediamine

(CH2)4

H 2(n − 1)H2O + H

COOH

Adipic acid

[

(CH2)6

O (CH2)4

Nylon 66

Reaction Mechanism (i) In polymerisation, the monomers should have a minimum of two reactive functional groups for polymerisation. (ii) Polymerisation proceeds in a step-wise reaction between reactive functional groups. In this process, the first two monomer units condense to form the dimer. The dimer reacts with another dimer to form a tetramer or with monomer to form a trimer. In this process, small units such as NH3, H2O, HCl and so on, are also formed. (iii) Only one type of reaction between two functional groups is involved in polymer formation. (iv) The polymer formed still contains both the reactive functional groups at the end of the chain; hence it is active and not dead as in chain polymerisation. (v) Reaction between two monomers contains two active functional groups. They will give straight chain polymers; otherwise, monomers containing more than two functional groups form crosslinked polymer. Formation of Polyester O nHO

R Diol

OH + nHOOC

COOH

Dicarboxylic acid

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O R

+ 2(n − 1)H2O

Polyester

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Polymers HO

OH + HOOC −H2O

R HO

COOH

OOC

2.7

COOH

Ester

HO HO

OOC

OH HOOC

COO

COOH

HOOC R COOH COO R COO R COO HOOC Triester

HOOC

HO HO

OOC

COO

HOOC

Diester

OH COO

R R

R OH

COO R COO R OH COO R COO R COO

HOOC

COO

HOOC

Triester

COOH

OCO

COO

COOH

R OOC

COOH

The process is continuous and forms polymers. Formation of Polyethylene Terephthalate CH2

CH2

OH + HOOC

Ethylene glycol HO CH2

CH2

COOH

HOOC HO

CH2

CH2 Ester

CH2

OOC Diester

COO

CH2

OOC

COO

COOH

Teryphthalic acid −H2O

Triester

CH2

COO

CH2 CH2 Diester

COO

CH2 Triester

COOH

OOC

Triester HO CH2

COOH

OOC

HOOC

COOH

HOOC

CH2 HOOC

CH2

COOH

OOC

CH2

COO

OH CH2

CH2

The aforementioned reaction is continued and forms a large polymer. n molecules of ethylene glycol react with n molecules of terephthalic acid. nHO

CH2

HO [ CH2

OH + nHOOC CH2

COOH COO

OOC

H + 2(n − 1)H2O

PET

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2.8 Engineering Chemistry Formation of Nylon 66 (CH2)4 H2N (CH2)6 NH2 + HOOC Hexamethylene diamine Adipic acid

COOH

−H2O

H2N (CH2)4

HOOC

COOH

(CH2)6

H2N

(CH2)6

(CH2)4

Diamide

(CH2)6

(CH2)4

HOOC

(CH2)6

OC (CH2)4 CO NH Triamide (CH2)4 HOOC

COOH

NH2

CO NH

(CH2)6 NH NH Diamide OC (CH2)4 COOH

NH (CH2)6 Triamide

COOH

(CH2)4

NH2 HOOC

(CH2)6

H2N

(CH2)4

OC NH Amide

(CH2)6

H2N

(CH2)4

COOH

H2N

(CH2)6

NH2

(CH2)6

NH2

The aforementioned reaction proceeds that n molecules of hexamethylenediamine react with n molecules of adipic acid to form nylon 66. nH2N

NH2 + nHOOC

(CH2)6 H

[ NH

(CH2)6

(CH2)4 NH

COOH (CH2)4

]n OH + 2(n − 1)H2O

Nylon 66

Formation of Polyamide H2N

NH2 + HOOC

H2N H 2N

H 2N HOOC

R H2N

NH R Diamide

R CO HOOC

NH OC Amide

NH2

HN R R CO

HOOC

NH2 NH

COOH R

COOH

R NH Diamide

COOH H2N

NH OC Triamide

R NH HOOC R

OC CO

R COOH NH R NH

NH2

Triamide

n molecules of diamine and n molecules of dicarboxylic acid react to form polyamide. nH2N

R Amide

NH2 + nHOOC

COOH

Dicarboxylic acid

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[ NH

] n OH + 2(n − 1)H2O

Polyamide

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Polymers

2.9

2.4.2 Addition/Vinyl/Chain Polymerisation Addition polymerisation takes place in compounds containing reactive double bonds. Chain polymerisation is characterised by a self-addition of the monomer molecules to each other very rapidly through a chain reaction. No by-product, such as HCl, NH3, H2O and so on, is formed. This polymerisation occurs in the presence of catalyst, light, or heat. (i) For example, In the presence of oxygen or Ziegler–Natta catalyst, ethylene gives addition polymer, that is, polythene nH2C

O2 (or)

CH2

Ziegler-natta catalyst

Ethylene

[ CH2

CH2

Polythene

(ii) In the presence of azo compounds, vinyl chloride gives polyvinyl chloride. nH2C

Azo compounds

[ H2C

CH ]n Cl

Vinyl chloride

Polyvinyl chloride

(iii) In the presence of metal, amide styrene gives polystyrene. nH2C

Metal amide

[ H2C

C6H5 Styrene

C6H5 (Polystyrene)

(iv) In the presence of benzoyl peroxide, acrylonitrile gives polyacrylonitrile. nH2C

Benzoyl peroxide

CN Acrylonitrile

[ H2C

CN Polyacrylonitrile

Reaction Mechanism In the addition polymerisation, free radical, carbonium ion, or carbanium ions, act as active centres. Hence, polymerisation may occur in the following: (i) Free radical (ii) Ionic (cationic or anionic) (iii) Coordination mechanisms In addition, polymerisation, consists of the following steps: (i) Chain initiation (ii) Chain propagation (iii) Chain termination Free Radical Polymerisation The initiation of the polymer chain is brought about by free radicals produced by the decomposition of monomers; thus, this reaction is called free radical polymerisation.

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2.10 Engineering Chemistry The decomposition of the initiation to form free radicals can be induced by heat energy, light energy or catalysts. Three steps included in free radical polymerisation are as follows:

(i) Chain initiation: (a) In the presence of catalyst, heat or light, the initiator converts into free radical. •

∆/ hv/catalyst

Initiator

Normally, H2O2, benzoyl peroxide, hydroperoxide, tertiary butyl peroxide and azobis-isobutyl nitriles (AIBN) act as initiators. (b) Free radicals react with the monomer unit to form a new, active-centred free radical. •

R + H2C Free radical

•

H2C

Monomer

New free radical

(ii) Chain propagation: (a) In this step, free radical attacks the fresh monomer unit to form another free radical. R

•

CH2

CH + H2C

CH2

•

CH2

New free radical

(b) This process continues till the monomer units are present in the reaction mixture. Finally, it forms a chain propagate free radical. CH2

•

CH + nCH2

Free radical

Monomer

CH2

[ CH2

•

]n CH2

Chain propagate free radical

Propagation reaction is a very fast reaction; in this reaction, no middle product is formed. (iii) Chain termination: In this step, chain propagate polymer radical deactivates with coupling or disproportionation reaction to stop chain propagation and forms dead polymer. (a) Coupling: Chain propagation free radicals meet to form polymer and terminate the reaction. R

CH2

•

CH + CH CH2

CH2

Polymer

(b) Disproportionation reaction: Hydrogen atom shifts from one chain propagate free radical to another chain propagate free radical and ends the reaction. R

CH2

•

CH + HC

CH2

CH2 CH2 + HC X

For example, polymerisation of acrylonitrile in the presence of benzoyl peroxide.

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Polymers

2.11

Reaction Mechanism

(i) Chain initiation: (a) In this step, benzoyl peroxide dissociates to form phenyl free radicals. C6H5

•

C6H5

•

2C6H5 + 2CO2

2C6H5COO

Benzoyl peroxide

Phenyl free radical

(b) Phenyl free radicals react with acrylonitrile and gives chain initiate free radical. •

R + H2C

CH2

Acrylonitrile

•

Chain initiate free radical

(ii) Chain propagation: (a) In this step, free radicals react with monomer units and give a chain propagate free radical. This reaction continues till the monomer units are present in reaction mixture. R

CH2

•

CH + HC CN

CH2

•

CH2

nH2C

CH CN •

CH ]n CH2

R [ CH2

CH CN

Addition polymer radical

(iii) Chain termination: (a) The chain terminates with the coupling reaction. R

CH2

•

CH + HC CN

CH2

CH CN

CH2

Addition polymer polyacrylonitrile (PAN) (or) (orlon)

For example, polymerisation of methyl methacrylate in the presence of azobis-isobutyl nitrile. Reaction Mechanism

(i) Chain initiation: (a) In the presence of ultraviolet rays, azobis-isobutyl nitrile homolysis gives cyanopropyl radical. (CH3)2

C CN

Azobis-isobutyl nitrile

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(CH3)2

U.V

2 (CH3)2

•

C + N2 CN

Cyanopropyl radical

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2.12 Engineering Chemistry

(b) Cyanopropyl radicals react with methyl methacrylate monomer to form chain initiate free radical. CH3

CH3 (CH3)2

•

C + H2C

COOCH3

(H3C)2

Cyanopropyl radical

C•

CH2

Methyl methacrylate

COOCH3

Chain initiate free radical

(ii) Chain propagation: (a) In this step, the chain initiates free radicals that react with monomer units; finally, it gives the chain propagate free radical. CH3 CH3 CH3 CH3

(H3C)2

C CN

CH2

C• + H2C

(H3C)2

COOCH3

CH2

C• COOCH3

CH3 C

nH2C

(H3C)2

COOCH3

CH3

C [ CH2

C ]n CH2

C•

COOCH3 COOCH3 Chain propagate free radical

(iii) Chain termination: (a) It terminates with a coupling reaction. CH3 CH3 CH2

C•

•

CH2

H2C

CH3

CH2

COOCH3 COOCH3

Polymethyl methocrylate (PMMA)

CH3

CH3 nH2C

ABIN

[ H2C

COOCH3 Methyl methocrylate

COOCH3 PMMA

Cationic Addition Polymerisation The following are the important points in reaction mechanism: (i) In this polymerisation, carbonium ion acts as the chain initiator.

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Polymers

2.13

(ii) BF3, H2SO4 and AlCl3 like Lewis acids act as catalyst, water, carbonic acid, etc., act as co-catalysts. (iii) In this polymerisation monomer acts as an electron donor and reacts with catalysts (electron acceptor) to form carbonium ion. The formed carbonium ion reacts with monomer units and propagates the reaction. For example, cationic polymerisation of isobutylene in the presence of BF3: CH3

CH3 nH2C

BF3

Isobutylene

C ]n

[ H2C

H2O

CH3

Polyisobutylene

Reaction Mechanism

(i) Chain initiation: (a) Carbonium ion is formed due to the following reactions. BF3 + H Catalyst

−

[BF3OH]H

Co-catalyst

CH3

CH3 [BF3OH]−

+ H + H2C

H3C

− [BF3OH]

C+ CH3

CH3

H+

[BF3OH]

Carbonium ion

(ii) Chain propagation: (a) Monomer molecules react one by one with carbonium ion and propagate the chain to form an addition polymer. CH3 CH3 CH3 CH3 H3C

C+

− [BF3OH] + H2C

CH3

H3C

−

[BF3OH]

CH3

Carbonium ion

C+

CH2

Isobutylene

CH3 C

nH2C

CH3

H3C

CH3

C [ CH2

C ]n CH2

C+

CH3

−

[BF3OH]

Addition polymer ion

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2.14 Engineering Chemistry

(iii) Chain termination: (a) Chain termination occurs with disproportionation reaction. CH3 H3C

[ CH2

CH3

C ]n CH2

C⊕

CH3

CH3 −

[BF3OH]

H3C

[ CH2

CH3

CH2

C ]n CH2

C + [BF3OH]H

CH3

Polyisobutylene

Anionic Addition Polymerisation The following are the important points: (i) (ii) (iii) (iv)

Carbanion acts as chain initiator. Alkali metal amide, Grignard reagents, etc., act as catalysts and initiate the chain. Monomers containing electron-withdrawing groups participate in anionic polymerisation. In this polymerisation, there is no chain termination; hence anionic polymers are known as activated polymers.

Acrylonitrile, styrene and methyl methacrylate participate in anionic polymerisation. For example, Anionic polymerisation of styrene in the presence of alkali metal amide: CH2

KNH2

[

CH2

]n Polystyrene

Styrene

Reaction Mechanism

(i) Chain initiation: (a) Carbanion is formed in the chain initiation. KNH2

(−)

K++ NH2

Catalyst (−)

NH2 + H2C

NH2

CH2

Carbanion

−

CH C6H5

(ii) Chain propagation: (a) Carbanion reacts with styrene monomers and propagates the reaction and forms chain propagate polymer ion. (b) In this type of polymerisation, chain termination does not occur. Hence, anionic polymer is active till the monomer units are present in the reaction mixture; therefore, anionic polymers are also known as active polymers.

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Polymers

NH2

CH2

−

CH + H2C

C6H5

NH2

CH2

−

CH2

C6H5 nH2C H2N [ CH2

2.15

C6H5 C6H5

CH −

CH ]nCH2

C6H5

Anionic polymer ion

(c) When impurities are added to the reaction mixture, chain termination occurs. Therefore, chain termination reagent is added and gets dead polymers. Differences between addition polymers and condensation polymers Addition polymer

Condensation polymer

1. Generally, vinyl monomers participate in addition polymerisation. 2. Reaction initiates with the help of an initiator. Initiator reacts with electrons of vinyl monomers and initiates the reaction. 3. In this process, chain initiation, chain propagation and chain termination steps are involved. Reactions are very fast; hence polymer is formed quickly. 4. Here HCl, H2O, NH3, etc., are not formed. 5. Monomer units always react with polymer radical or polymer ion. 6. For example, polyethylene, polyvinyl chloride, Teflon, etc.,

Monomers contain minimum two active functional groups participate in condensation polymerisation Here, there is no need of initiator and reaction occurs between monomer units. Step-wise reaction occurs in between functional groups; reaction is slow. Finally, it gives polymer that contains high molecular weight. Small units like HCl, H2O, NH3, etc., are formed. Monomer reacts with another monomer units or polymer molecule. For example, nylon 66, polyester, etc.,

2.4.3 Coordination Polymerisation Coordination polymerisation is invented by two Italian scientists, Ziegler and Natta. They shared the Nobel Prize for Chemistry in 1963 using the Ziegler–Natta catalyst to polymerise non-polar monomers. Ziegler–Natta Catalyst The mixture of titanium halides and trialkyl aluminium is known as Ziegler–Natta catalyst. (TiCl3 + R3Al) The reaction between titanium chloride and trialkyl aluminium forms Ziegler–Natta catalyst. In this process, trialkyl aluminium adsorbs on the surface of titanium chloride, and forms electron deficient bridge structure.

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2.16 Engineering Chemistry R R R

Cl Cl

Ti Cl

Ziegler-Natta structure

In this structure, titanium chloride acts as catalyst and trialkyl aluminium acts as the co-catalyst. Coordination polymerisation is a form of addition polymerisation in which monomer adds to a growing macro molecule through an organometallic active centre. Reaction Mechanism (i) Cossee–Arlman mechanism: The formation of C C bonds in the polymerisation of alkenes are explained by Cossee–Arlman. In this pathway, an intermediate coordination complex contains both growing polymer chain and the monomer. The coordination polymerisation of alkene can be proceeded by monometallic or bimetallic mechanism depending on the catalyst. Alkene or substituted alkene is polymerised by Ziegler– Natta catalyst. The double bond of alkene will undergo cis addition with the empty orbital of Titanium catalyst to form a four-membered ring coordinate intermediate. (ii) Monometallic polymerisation mechanism: R

Cl Cl

R Ti

Cl Cl

R Ti

CH R1

R1 Cl

CH2 Ti

Four membered ring coordinate intermediate

(iii) Bimetallic polymerisation mechanism: If the catalyst is made from aluminium and titanium compounds, the polymerisation proceeds through bimetallic mechanism. In this mechanism, first, a bridge structure is formed between two metal compounds. Then, the substituted alkene is coordinated with titanium compound to form a six-membered ring. R R

R R

H2C

δ+

Ti δ−

R′

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R′

δ−

δ+

CH2 R

CH Al

CH R

− δ+ δ CH2

R′ Ti

R′

CH2 Al R

CH R′

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Polymers

2.17

Importance The Ziegler–Natta catalyst can control the linearity and tacticity of the polymer. Free radical polymerisation of ethylene produces a low density branched chain soft, rubbery polymer. However, the Ziegler–Natta catalyst produces more linear, rigid, high density, high tensile strength, harder and tougher isotactic polymer. In the presence of the Ziegler–Natta catalyst, coordination polymerisation occurs and gives isotactic polymer of olefin. For example, propylene undergoes coordination polymerisation in the presence of Ziegler–Natta catalyst at 50°C and gives isotactic polymer of polypropylene. nCH

CH2

TiCl3 + R3Al 50°C

H2C

CH3

H2C

CH3

Propylene

CH CH3

H2C

CH3

Polypropylene

2.5 MOLECULAR MASS OF A POLYMER Polymers are a mixture of different monomers with different molecular weights or masses. Hence, three kinds of molecular masses have been identified, which are as follows: (i) Number average molecular mass ( M n ) : This is the total mass (w) of all the molecules in a polymer sample divided by the total number of molecules present. This can be determined by measuring colligative properties like freezing point depression, boiling point elevation, osmotic pressure, lowering of vapour pressure, etc. Mn =

w = ∑N

∑ Ni Mi ∑ Ni

Where Ni = the number of molecules of mass Mi The number average molecular mass is a good index for tensile strength, but not for flow. (ii) Weight average molecular mass ( M w ) : This can be determined from light scattering and ultracentrifugation techniques and can measure molecular size as follows: Mw = where wi = weight fraction of molecules of Mi Mw =

∑ wi M i ∑ wi

∑ Ci M i = ∑ Ci M i = ∑ N i M i2 C ∑ Ci ∑ Ni Mi

where Ci = weight concentration of Mi molecules C = total weight concentration of all polymer molecules Polydispersity index or molecular mass distribution This is a measure of the distribution of molecular mass of a polymer. This can be calculated using the weight average molecular weight divided by the number average molecular weight. Polydisperity index =

M02_ENGINEERING-CHE00_SE_1181_CH02.indd 17

Mw MN

5/21/2016 9:20:49 AM

2.18 Engineering Chemistry In a monodisperse system, M w = M N . However, the polydispersity index (PDI) value is always greater than one, that is, the weight of the average molecular mass is always greater than the number average molecular mass. (iii) Viscosity average molecular mass ( M v ) : Viscosity average molecular mass can be determined by the measuring of viscosity of that particular polymer. This can be explained by the following formula: 1

 ∑ wi M ia   ∑ N i M ia +1  a Mv =   =  ∑ wi   ∑ wi M i 

where a = constant. When a = unity, the viscosity and weight average molecular masses are equal. M v is almost less than M w , a polydispersive polymer is represented as Mw > Mv > Mn .

2.6 PLASTICS Plastics are mainly of two types: (i) Thermoplastics: Thermoplastics are formed with straight chains; hence when heated, they soften, subsequently melt and when cooled, they become hard. Due to van der walls forces, two different chains slid over each other and the plasticity of such polymers is reversible. Hence, the materials can be moulded and remoulded without damage. For example, PVC, nylon, polystyrene and polyethylene (ii) Thermosetting plastics: Thermosetting resins do not become soft on heating, and they never melt once they set. They are normally made from semi-fluid polymers with low molecular masses. In this case, the polymer chains are entangled with one another and hence cannot slide over each other. Deformation does not occur on heating, because only the primary covalent bonds are present in the entire structure, forming a three dimensional network. They have no scrap value. For example, Bakelite Differences Between Thermoplastics and Thermosetting Plastics Thermoplastics

Thermosetting plastics

1. They soften on heating and harden on cooling.

They do not soften on heating; on prolong heating, however, they burn. These plastics have three dimensional network structure, joined by strong covalent bonds. They are formed by condensation polymerisation. They retain their shape and structure, even on heating. Hence, they cannot be reshaped and reused. They are usually hard, strong and more brittle. They cannot be reclaimed from wastes. Due to strong bonds and cross-linking, they are insoluble in almost all organic solvents. For example, Bakelite, novalac resin, resol, resin, etc.

2. They consist of long-chain linear macromolecules. 3. They are formed by addition polymerisation. 4. By reheating to a suitable temperature, they can be softened, reshaped and thus reused. 5. They are usually soft, weak and less brittle. 6. They can be reclaimed from wastes. 7. They are usually soluble in some organic solvents. 8. For example, polyethylene, polyvinyl chloride, polystyrene, etc.

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Polymers

2.19

2.7 IMPORTANT POLYMERS—COMPOSITION, PREPARATION, PROPERTIES AND ENGINEERING USES 2.7.1 Thermoplastics Polythene or Vinyl Resin Polythene is the most widely used plastic. Polythene is obtained by high-pressure polymerisation of ethylene, making use of oxygen as initiator. The reaction takes place at 1,500 atmospheric pressure and 180°C–250°C temperature range. Ethylene polymerised into a waxy solid known as polyethylene. H

n C

Polymerization

Polyethylene

Ethylene

By using free radical initiator, low density polythene (LDPE) is obtained, while by using ionic catalysts, high density polythene is obtained. Properties The properties of polythene are as follows: (i) Polythene is a rigid, waxy, white, translucent non-polar material, exhibiting considerable chemical resistance to strong acids, alkalis and salt solutions at room temperature. (ii) It is a good insulator of electricity. However, it is swollen and permeable to most oils and organic solvents, particularly kerosene. (iii) Due to its highly symmetrical chain structure, polythene crystallises very easily. The degree of crystallinity may vary from 40%–95%, depending on the degree of branching in the polythene chain. (iv) Polyethylene produced by high-pressure process has a branched structure and therefore, is flexible and tough. On the other hand, the low-pressure process results in a completely linear polyethylene, having higher density and better chemical resistance. (v) Commercial polythene can be subdivided into three groups: (a) Low-density polythene (LDPE): It is polymerised under very high pressure of 1,000–5,000 atmospheres and temperature range of 80°C–250°C in the presence of O2 as the initiator. It has a density of 0.91–0.925 g/cm3 and M.P. 110°C –125°C. (b) Medium-density polyethylene (MDPE): It is polymerised under medium pressure, having density 0.925–0.940 g/cm3 and M.P. 130°C–140°C. (c) High-density polyethylene (HDPE): It is polymerised under atmospheric pressure (6–7 atmospheres) and temperature at 60°C in the presence of Ziegler–Natta catalyst [TiCl3 + Al(C2H5)3]. The HDPE, which is completely linear, has better chemical resistance and higher softening point but relatively brittle. Comparative properties of HDPE and LDPE Type

Density (g-cm−3)

Crystallinity

Softening point (k)

Tensile strength (atm)

LDPE HDPE

0.910–0.925 0.941–0.965

55% 80%

360 400

85–136 204–313

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2.20 Engineering Chemistry Uses The uses of LDPE are as follows: (i) LDPE is used in making films and sheets. Pipes made of LDPE are used for agricultural irrigation and domestic water line connections. They are also used for making tubes, coated wires and cable wires. (ii) HDPE is used in the manufacture of toys, insulator parts, bottle caps, flexible bottles, kitchen and domestic articles. Polyvinyl Chloride (PVC) It is obtained by heating a water emulsion of vinyl chloride in the presence of a small amount of benzoyl peroxide or hydrogen peroxide in an autoclave under pressure. H

n C

Polymerization

Vinyl chloride

PVC

Vinyl chloride so needed is generally prepared by treating acetylene at 1–1.5 atmospheres with hydrogen chloride at 60–80°C, in the presence of metal chloride as catalyst. CH2 CHCl CH CH + HCl Acetylene

Vinyl chloride

Properties PVC is colourless, odourless, non-inflammable and chemically inert powder, resistant to light, atmospheric oxygen, inorganic acids and alkalis but soluble in hot chlorinated hydrocarbons such as ethyl chloride. Pure resin possesses high softening point and a greater stiffness and rigidity, but is brittle. Uses The uses of PVC are as follows: (i) Rigid PVC or unplasticised PVC has superior chemical resistance and high rigidity but is brittle. It is used for making sheets, which are employed for tank linings, light fittings, safety helmets, refrigerator components, tyres, cycles and motor-cycle mudguards. It is also extruded in strip and tube form for use in place of nonferrous metals. (ii) Plasticised PVC is used for making continuous sheets, rain coats, table cloths and curtains, electric cables, toys, radio components, conveyor belts, etc. Polystyrene (PS) It is prepared by the polymerisation of styrene in the presence of benzoyl peroxide catalyst. H

n C

C H

Styrene

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Polymerization

C H

Polystyrene

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Polymers

2.21

Properties Polystyrene is a transparent, light-stable and moisture-resistant material. It is highly electric insulating and highly resistant to acids, and is a good chemical resistant. However, it has less softening and is brittle. It has the unique property of transmitting light through curved sections. Uses It is used in moulding of articles such as toys, combs, buttons, buckles, radio and television parts, refrigerator parts, battery cases, high-frequency electric insulators, lenses, indoor lighting panels, food containers, food packaging, umbrella handles and so on. Teflon [Polytetrafluoroethylene (PTFE or FLUON)] The presence of benzoyl peroxide catalyst and high pressure polymerisation of tetrafluoroethylene gives Teflon. F n

Tetrafluoroethylene

Polymerization

Teflon

Properties Teflon has a twisted, zigzag structure with fluorine atoms, packed tightly in a spiral around the carboncarbon skeleton. Due to the presence of highly electronegative fluorine atoms, there are very strong attractive forces between different chains. These strong attractive forces give the material extreme toughness, high softening point, exceptionally high chemical resistance towards all chemicals, high density, waxy touch, very low coefficient of friction and extremely good electrical and mechanical properties. It can be machined, punched and drilled. The material cannot be dissolved and cannot exist in a true molten state. Around 350°C, it sinters to form a very viscous, opaque mass, which can be moulded by applying high pressure. Uses The uses of Teflon are as follows: (i) It is used as an insulating material (motors, transformers, cables, wires, etc.) and for making gaskets, pickings, pump parts, tank lining, chemical-carrying pipes, tubings and tanks, etc. (ii) It is used for coating and impregnating glass fibres, asbestos fibres, cloths, in non-lubricating bearings, non-sticking stop-clocks, coating of frying pans, etc. Nylon These are polyamides. The word nylon is now accepted as a generic term for synthetic polyamides, which are characterised by a repeating acids linkage ( NHCO ). Nylon is formed with dicarboxylic acids and diamide under condensation process. It has been named on the basis of number of carbon atoms present in that two monomer units.

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2.22 Engineering Chemistry For example, nylon 6,6, nylon 6,10, nylon 6,11, etc. Nylon 6,6 is formed with the condensation reaction of hexamethylenediamine and adipic acid. H

H HO +

N H

C O

Hexamethylene diamine

C H

OH C 4

Condensation Polymerization −H2O

Adipic acid

+ 2nH2O

C 4

Nylon 6,6

Properties They are translucent, whitish, horny and high melting polymers. They possess stability up to high temperature and good abrasion resistance. They are insoluble in common organic solvents and soluble in phenol and formic acid. Properties of Nylon Fibres The properties of nylon fibres are as follows: (i) (ii) (iii) (iv) (v) (vi) (vii)

They are light, horny and high melting. They are insoluble in common solvents. They have good strength. They absorb little moisture and are thus “drip-dry” in nature. They are very flexible and retain original shape easily. They have resistances to abrasion. On blending with wool, the strength and abrasion resistance of the latter increase.

Uses (i) Nylon 6,6 is primarily used for fibres and tyre cord, which find use in making rocks, ladies hoses, undergarments, dresses, carpets, ropes, etc. (ii) Nylon 6,6 and nylon 6,11 are mainly used for moulding purposes for gears, bearings, electrical mountings, etc. Nylon bearings and gears work quietly without any lubrication. (iii) They are also used for making filaments for ropes, bristles for tooth brushes, films, etc.

2.7.2 Thermosetting Plastics Phenol-Formaldehyde Resins or Phenoplasts These are condensation polymerisation products of phenolic derivatives with aldehydes, prepared by condensing phenol with formaldehyde in presence of acidic or alkaline catalyst. Depending upon catalyst and reactants mainly three kinds of resins are formed, they are classified as follows: (i) Novalac resin: The presence of acid, phenol and formaldehyde condense to form novalac resin. At first, formaldehyde takes proton from acid and forms the carbonium ion. H H C + H+ H2C+ OH O Formaldehyde

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Carbonium ion

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Polymers

2.23

Phenol react with carbonium ion to form ortho- and para-methylol phenol. OH

OH CH2OH

+ H2C

+ CH2OH

Phenol

O-methylol phenol

P-methylol phenol

Ortho-methylol phenol condenses to form novalac resin. OH

CH2

n OH

CH2OH

+ OH

CH2

OH CH2

CH2

Novalac resin

(ii) Resol resin: Phenol and formaldehyde are refluxed with ammonia at 100°C gives resol resin. In presence of ammonia, methylol phenol has greater reactivity with formaldehyde than phenol, hence it gives di- and tri-methylol products. OH CH2OH

HOH2C

CH2OH

HCHO

OH CH2OH

HCHO

CH2OH Mono-methylol phenol

Di-methylol phenol

Tri-methylol phenol

The polymethylol phenol condenses and forms resol resin. OH HO

H2C

OH CH2

CH2OH

OH CH2 n

CH2OH

OH CH2

CH2OH

Resol resin

(iii) Bakelite: Bakelite was first prepared by Baekeland. In the presence of hexamethylenetetramine, phenol reacts with formaldehyde and forms cross linked resin, that is, Bakelite. It is a hard and insoluble solid.

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2.24 Engineering Chemistry OH H

OH H

H H

(CH2)6N4 H2O

O H OH

CH2

OH CH2

CH2

Bakelite

Properties Novalac resin is soluble and fusible solid. Resol resin is a hard and brittle solid. Bakelite is a rigid, hard, infusible, water resistant, and insoluble solid. It resist to non-oxidising acids, salts and organic solvents, but are attacked by alkali due to presence of free hydroxyl groups. All phenol formaldehyde resins possess excellent electrical insulating character. Uses The uses of resol resin are as follows: (i) (ii) (iii) (iv) (v) (vi)

Making electric insulator parts like switches, plugs, switch-boards, heater-handles, etc. Preparing moulded articles like telephone parts, cabinets for radio and television. Impregnating fabrics, wood and paper. As adhesives for grinding wheels. Making bearings used in propeller shafts for paper industry and rolling mills. As hydrogen-exchanger resins in water softening.

Polyurethanes Diisocyanate and diol give polyurethanes. For example, a reaction between 1,4-butane diol and 1,6-hexane diisocyanate gives “Perlon - U”, a crystalline polymer. Properties The properties of polyurethanes are as follows: (i) Polyurethanes are less stable than nylons. (ii) These have excellent resistance to abrasion and solvents.

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Polymers

2.25

Uses These are used as coatings, films, foams, adhesives and elastomers. O nC

H N

C H

C + n HO

C H

OH 4

1,4-butane diol

1,6-Hexamethylene diisocyanate

O n

H O

C H

(Perlon-u)

2.8 RUBBER (ELASTOMERS) Rubbers are high polymers, which have elastic properties. Thus, the rubber band can be stretched to four to 10 times its original length, and as soon as the stretching force is released, it returns to its original length. The elastic deformation in an elastomer arises from the fact that in the unstressed condition, an elastomer molecule is not straight chained, but in the form of a coil, it can be stretched like a spring consequently. The unstretched rubber is amorphous.

2.8.1 Processing of Natural Rubber Isoprene is the basic molecule present in natural rubber. Dispersive form of isoprene units are known as latex. In the processing of natural rubber isoprene molecules polymerise and form long, coiled chains of cis-polyisoprene. H2C n

C H3C

CH2 C

H H

Isoprene (2-methyl 1,3-butadiene)

Cis-polyisoprene (Natural rubber)

H C H H H C H H H C H H H C H H H H H H H H H H C C C C C C C C C C C C C C C C H H H H H H H H Structure of natural rubber

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2.26 Engineering Chemistry By making small incisions on the bark of rubber trees, like having a brasiliensis and guayule, the rubber latex can be collected into small vessels, as it oozes out. It contains 25%–45% of rubber in the form of milky colloidal emulsion, the remainder of which is made mainly of water and small amounts of protein and resinous material with time, the flow of latex from the incision made start decreases. Thus, at regular intervals, tapping is necessary throughout the life of the tree. Latex is diluted to make 15%–20% of rubber and is filtered to eliminate any dirt present in it. It is then coagulated in a tank, fitted with irregular partitions by adding about one kg of acetic acid or formic acid per 200 kg of rubber, to a soft white mass. After washing and drying, the coagulated residue is treated as follows: (i) Crepe rubber: It is prepared by adding a little sodium bisulphite to bleach the rubber and is then passed through a creping machine so that coagulum is rolled out into sheets of about 1 mm thickness. The sheet possesses an even rough surface resembling crepe paper. The sheet is then air-dried at 50°C. (ii) Smoked rubber: It is made by eliminating the bleaching with sodium sulphite and rolling the coagulum into thicker sheets, having ribbed pattern on its surface. The ribbed surface pattern on the sheet prevents them from adhering together on stacking. It also facilitates consequent drying as it exposes greater surface area of the sheet. The sheets are then dried in the smoke obtained from burning wood or coconut shells at about 50°C. The rubber sheet thus obtained is translucent and amber in colour.

2.8.2 Gutta–Percha It is a trans-form of natural rubber. (In natural rubber, isoprene units are linked with cis-form). It is obtained from the matured leaves of Dichopsis gutta and Palaquium gutta trees, grown mostly in Malaya and Sumatra. Gutta–percha can be recovered by solvent extraction process, when insoluble resins and gums are separated. Alternatively, the matured leaves are grounded carefully and is treated with water at about 70°C for half an hour and then poured into cold water, when Gutta–percha floats on water surface it is removed. −H2C H3C

H3C

CH2

H3C

CH2

CH2 H

Rod structure (Gutta percha)

Properties The properties of gutta percha are as follows: (i) At room temperature, gutta percha is horny and tough, but softens and becomes tacky at about 100°C. (ii) It is soluble in aliphatic hydrocarbons, but insoluble in aromatic and chlorinated hydrocarbons. Uses It is used in the manufacturing of golf ball covers, submarine cables, adhesives and tissues for surgical purposes.

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Polymers

2.27

2.8.3 Vulcanisation of Rubber Drawbacks of Raw (Natural) Rubber The drawbacks of raw (natural) rubber are as follows: (i) It is plastic in nature. Crude rubber becomes soft and sticky in hot summer, while in cold weather, it becomes hard and brittle. Its usage is limited to a particular temperature range of about 10°C–60°C. (ii) It is weak, due to low tensile strength (200 kg/cm2). (iii) It has large water absorption capacity. (iv) It is non-resistant to non-polar solvents such as vegetable and mineral oils, gasoline, benzene and carbon tetra chloride. (v) It is attacked by oxidising agents (HNO3, H2SO4, etc.). It perishes due to the oxidisation in air. (vi) In organic solvents, it undergoes swelling and gradual disintegration. (vii) It possesses tackiness, which means that under pressure, two fresh raw rubber surfaces coalesce to form a single piece. (viii) It possesses very less durability. (ix) When stretched to a great extent, some molecular chains undergo sliding or slipping over each other, hence it suffers from permanent deformation. To improve the properties of rubber, it is compounded with some chemicals such as sulphur, hydrogen sulphide, benzoyl chloride and the rubber mix is prepared for vulcanisation. The addition of compounding agents is facilitated by the process of mastication. Mastication of rubber means that it is subjected to severe mechanical working. Oxidative degradation accompanied by a marked decrease in the molecular weight of the rubber occurs and converts rubber into a soft and gummy mass. Vulcanisation Heating of raw rubber with sulphur at around 100°C –140°C is known as vulcanisation. Sulphur combines chemically at the double bonds of rubber chains and forms cross links. With these crosslinks, the rubber becomes stiff and the percentage of sulphur determines the stiffness of rubber. For example, a rubber tyre contains 3–5% of sulphur. CH3 CH3 CH2

CH2

CH3

CH2

C CH3

Raw rubber Vulcanization addition of sulphur

CH3 CH2

CH2

CH3

CH2

CH3

Sulphur Cross-links

CH2

CH3

Vulcanization of raw rubber

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2.28 Engineering Chemistry Advantages of Vulcanisation The advantages of vulcanisation are as follows: (i) The tensile strength of vulcanised rubber is very good and has extensibility about 10 times the tensile strength of raw rubber; when a tensile force is applied, it can bear a load of 2000 kg/cm 2 before it breaks. (ii) After the removal of deforming force, the articles made from vulcanised rubber regain their original shape, so that it has excellent resilience. (iii) It has broader useful temperature range (−40°C–100°C) compared to raw rubber’s temperature range (10°C–60°C). (iv) It has much high resistance to moisture, oxidation and abrasion. (v) It has much higher resistance to wear and tear as compared to raw rubber. (vi) It is a better electrical insulator, although it tends to absorb small amounts of water. or example, ebonite is a better insulator. (vii) It has resistance to organic solvents like benzene, carbon tetrachloride, petrol, etc., but it swells in them. (viii) It is very easy to manipulate vulcanised rubber to produce the desired shapes. The superior properties of vulcanised rubber compared to raw rubber are summarised in Table 2.1. Table 2.1 Raw rubber vs. vulcanised rubber Property

Raw rubber

Tensile strength Resilience Resistance to moisture oxidation and abrasion Useful temperature range Resistance to organic solvents Tackiness Elasticity Per cent elongation at break Manipulability to desired shape

200 kg/cm Good Poor 10°C –60°C Poor Marked Very high 1200 Not easy

Vulcanised rubber 2,000 kg/cm2 Very good Good −40°C –100°C Large but limited Slight Low 800 Easy

2.8.4 Compounding of Rubber Compounding is “mixing of the raw rubber with other chemicals so as to impart the product-specific properties suitable for particular job”. The following substances are generally mixed with raw rubber: (i) Antioxidant: Natural rubber has a tendency to “perish” due to oxidation, retarding the deteriorating of rubber by preventing its oxidation by light and air; about 1% of the antioxidants are used. These are phenolic substances, phosphites and complex organic amines like phenyl naphthylamine. (ii) Colouring agents: Colouring agents impart the desired colour to the rubber product. For example, for white products, titanium dioxide (TiO2) is the usual pigment. For colour products, the following pigments are used:

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Colour

Pigment

Blue Green Red Crimson Yellow

Ultramarine Chromium trioxide Ferric oxide Antimony sulfide Led chromate

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Polymers

2.29

(iii) Vulcanising agents: The main vulcanising agent is sulphur. Among these, sulphur monochloride, hydrogen sulphide and benzoyl chloride are also used as vulcanisers. Depending on the nature of the product required, the percentage of S added varies between 0.15% and 32%. The process of vulcanisation brings about excellent changes in the properties of crude rubber. (iv) Accelerators: They are positive catalysts for vulcanisation process as they drastically shorten the time required for vulcanisation. Generally, benzothiazole, 2-mercaptol, zinc alkyl xanthate and thiocarbamates are used in 0.5%–1% instances as the most usual accelerators. (v) Reinforcing agents: These agents give rigidity, strength and toughness to the rubber and are present up to 35% of the rubber compound. Commonly used reinforcers are carbon black, zinc oxide, CaCO3, MgCO3, etc. For example, addition of carbon black in the elastomer is used in the manufacture of automobile tyres. (vi) Plasticisers and softness: They impart greater tenacity and adhesion to the raw rubber. The important plasticisers are stearic acid, waxes, vegetable oils, rosin, etc. (vii) Inert fillers: They lower the cost of the product and alter the physical properties of the mix for simplifying the subsequent manufacturing operations.

2.8.5 Synthetic Rubbers or Artificial Rubber The landmark discovery of rubber is the greatest achievement in polymer industry and with the efforts of scientists and technologists the first useful synthetic rubber Buna-S was prepared. Advantages of Synthetic Rubbers Due to better performance properties of synthetic elastomers, natural rubber failed to give stiff competition. (i) (ii) (iii) (iv) (v) (vi)

They are produced from petrochemical raw materials in abundant amounts. They are economically beneficial. They are not only replacements but are superior to natural rubber in certain cases. They are tailor-made elastomers with diverse applications. They have high abrasion resistance and high tensile strength. Certain elastomers like silicones have low temperature (−80°C), flexibility and high temperature stability. (vii) Silicones are valuable in surgical prosthetic devices.

2.8.6 Important Artificial Rubbers Styrene Butadiene Rubber (GR-S or Buna-S) It is a copolymer of styrene (25% by weight) and butadiene (75% of weight). The monomers are emulsified in water using soap or detergent. The reaction is initiated by peroxide initiators. Polymerisation is carried out at 5°C, and therefore, the product is known as cold SBR. nCH2

CH2 + nCH2

Polymerization

in presence of peroxide

C6H5 Butadiene (75%)

Styrene (25%)

( CH2

CH2

CH )n C6H5

Buna-S

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2.30 Engineering Chemistry Properties Styrene rubber is slightly inferior to natural rubber in its physical properties. It possesses high abrasion resistance, high load-bearing capacity and resilience. However, it gets readily oxidised, especially in the presence of traces of ozone present in the atmosphere. It swells in oils and solvents. It can be vulcanised in the same way as natural rubber, but it requires less sulphur and more accelerators for vulcanisation. Uses The uses of styrene rubber are as follows: (i) It is mainly used for the manufacture of motor tyres, (passenger car tyres, motor cycle tyres and scooter and cycle tyres.) but not used for truck tyres. (ii) It is also used for floor tiles, shoe soles, gaskets, footwear, conveyor belts, wire and cable insulations, carpet backing, adhesives, tank linings, etc. Buna-N, Nitrile Rubber or GR–A It is a copolymer of a 1,3-butadiene and acrylonitrile. They are also prepared in emulsion systems. They are noted for their oil resistance but not suitable for tyres. n [ H2C

CH2 ] + n (CH2

CH)

Polymerization

CN 1,3-Butadiene

Acrylonitrile

( CH2

CH2 )n ( CH2

CH )n CN

Nitrilerubber (Buna-N)

Properties It possesses excellent resistance to heat, sunlight, oils, acids and salts, but it is less resistant to alkalis than natural rubber because of the presence of cyano groups. As the proportion of acrylonitrile is increased, the resilience to acids, salts, oils, solvents, etc., increases, but the low temperature resilience suffers. Vulcanised rubber is more resistant to heat and ageing than natural rubbers and may be exposed to high temperature. Uses The uses are as follows: (i) They are extensively used for fuel tanks, gasoline hoses, creamery equipment, conveyor belts and high-altitude aircraft components. (ii) They are also used as adhesives, latex, gaskets, printing rollers, leather and textile. Poly Sulphide Rubber, Thiokol or GR–P Thiokols are those elastomers in which sulphur forms a part of the polymer chain. It is a copolymer of sodium poly sulphide (Na2S4) and ethylene dichloride.

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Polymers S CH2 CH2

Cl + Na

Ethylene dichloride

heated

CH2

−NaCl

Sodium polysulphide

2.31

Thiokol rubber

Properties Thiokols have outstanding resistance to swelling and disintegration by organic solvents, mineral oils, fuels, solvents, oxygen, ozone, gasoline and sunlight. Thiokol films have low permeability to gases. They have the following limitations: (i) It tends to flow or lose shaped under continued pressure. As it cannot be vulcanised, it does not form hard rubber. (ii) Its tensile strength is lesser than that of natural rubber. (iii) It cannot be vulcanised. Uses The uses of Thiokols are as follows: (i) Thiokols are used for barrage balloons, life rafts and jackets, which are inflated by CO2. (ii) It is also used for lining hoses for conveying gasoline and oil, in paints, for gaskets, diaphragm and seats in contact with solvents, and for printing rolls. Polyurethanes (Isocyanate Rubber) These polymers are formed by the reaction between diisocyanates and polyalcohols. O O nHO

CH2

OH + nC

Ethylene glycol

CH2

Polymerization

Ethylene diisocyanate

CH2

Polyurethane rubber (or) isocyanate rubber

Properties Polyurethane elastomers have outstanding abrasion resistance and hardness combined with good elasticity and resistance to oils, greases, chemical, weathering and solvents. Uses They are used in applications where extreme abrasion resistance is required such as in heel lifts, surface coatings, manufacture of foams, spandex fibres and small industrial wheels. Silicone Rubbers Silicones are organic silicone polymers. They have alternate Si-O-bonds. Preparation: For preparation of silicones, dialkyl-substituted silanes are used as raw materials.

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2.32 Engineering Chemistry They undergo hydrolysis and condensation polymerisation to form silicone polymers. Silicones are of mainly two types: (i) Straight-chain polymers: Dialkyl-substituted silanes are subjected to hydrolysis and undergo condensation polymerisation to form straight-chain thermoplastic silicones. R R2SiCl2

hydrolysis

R2Si(OH)2

condensation

Dialkyl silanol

Dialkyl dichloro silane

Si R

Straight chain thermo plastic silicone polymer

(ii) Cross-linked polymers: Mono alkyl-substituted silanes are subjected to hydrolysis and proceeds condensation reaction gives cross-linked polymers. R Cl

R Cl

hydrolysis

Si OH

Mono alkyl silanol

R nHO

condensation

OH Mono alkyl silanol

Si O

O O

Cross linked polymer

Alkyl chlorosilanes are prepared by this process. R

X + SiCl4

RSiCl3 + MgXCl Mono alkyl trichloro silane

R2SiCl2 + 2MgXCl Dialkyl dichloro silane

Properties The properties of silicones are as follows: (i) (ii) (iii) (iv)

Silicones have resistance to heat, cold, oxidation, etc. They obtain as liquids, resins and elastomers. Silicon liquids are having repulsion with water. They have also resistance with chemicals. Silicone rubber may form reaction between silicone resins, silica and alumina.

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Polymers

2.33

Uses The uses of silicones are as follows: (i) (ii) (iii) (iv)

Silicones are used as anti-forms in the preparation of food stuff. They are used as elastics, insulators, gascatoles and ceiling joints, and also used in medicine. They are used as surface-coating agents. Silicones are used for lamination.

Reclaimed Rubber Reclaimed rubber is rubber obtained from waste rubber articles such as worn-out tyres, tables, gaskets, hoses, foot wear and so on. The process of reclamation of rubber is carried out as follows: The waste is cut to small pieces and powdered by using a cracker, which exerts powerful grinding and tearing action. The ferrous impurities, if any are removed by the electromagnetic separator. The purified waste powdered rubber is then digested with caustic soda solution at about 200°C under pressure for 8–15 hours in a “steam-jacketed autoclave”. By this process, the fibres are hydrolysed. After the removal of fibres, reclaimed agents (like petroleum and coal tar based oils) and softeners are added and sulphur gets removed as sodium sulphide and rubber becomes devulcanised. The rubber is then thoroughly washed with water sprays and dried in hot air driers. Finally, the reclaimed rubber is mixed with small proportion of reinforcing agents (like clay, carbon black, etc.). Properties The reclaimed rubber is of less tensile strength lower in elasticity and possesses lesser wear-resistance than natural rubber. However, it is much cheaper, uniform in composition and has better ageing properties. Moreover, it is quite easy for fabrication. Uses For manufacturing tyres, tubes, automobile floor mats, belts, hoses, battery containers, mountings, shoes and heals, etc.

2.9 REINFORCED OR FILLED PLASTICS Polymers, generally of low strengths and moduli of elasticity, are needed for structural purposes. For these reasons, polymers are combined with fillers (which are primary silicates) to get better products. The fillers are solid additives, which modify the physical properties, particularly, the mechanical properties of basic polymeric materials. For example, they improve: (i) thermal stability, (ii) mechanical strength, (iii) insulating characteristics, (iv) water resistance, (v) external appearance, (vi) rigidity, (vii) finish, (viii) hardness, (ix) opacity and (x) workability, besides reducing (xi) cost and (xii) shrinkage on setting and brittleness. Usually, specific fillers are added to a polymeric compound to impart special characters to the final products. Some examples are as follows: (i) To improve hardness, fillers such as carborundum, quartz and mica are added. (ii) To make the plastic impervious to x-ray, barium salts are fillers. (iii) To provide heat and corrosion resistances, asbestos is added.

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2.34 Engineering Chemistry The combination of polymeric substance with solid fillers, is called filled or reinforced plastics. The filler acts as a reinforcing material while the polymer acts as binder, which links the filler particles. The polymer serves as stress transforming agent from filler to filler particles. Most commonly used fillers are as follows: (i) wood flour, (ii) saw dust, (iii) ground cork, (iv) asbestos, (v) marble flour, (vi) china clay, (vii) paper pulp, (viii) corn husk, (ix) mica, (x) pumice powder, (xi) carbon, (xii) cotton fibres, (xiii) boron fibres, (xiv) silicon carbide, (xv) silicon nitride, (xvi) graphite, (xvii) alumina, (xviii) glass fibres, (xix) Kevlar fibres, (xx) cotton fibres, (xxi) metallic oxides such as ZnO, PbO and so on, and (xxii) metallic powder such as Al, Cu, Pb and so on.

2.9.1 Composition Fillers are usually employed in a sizable weight percentage. The percentage of filler can be up to 50% of the total moulding mix.

2.9.2 Nature of Polymers Used Polymers used are thermoplastics, thermosetting polymers as well as rubber (elastomers) such as polyethylene, polypropylene, nylon-6, PET, polystyrene, melamine, silicone, natural and synthetic rubbers, epoxy among others. Some examples of filled plastics are as follows: (i) Carbon black on addition to natural rubber brings about 40% increase in tensile strength and such filled plastic is used in automobile tyres. (ii) The addition of china clay to PVC enhances the electrical insulation characteristics of PVC. Therefore, this filled plastic is used for electrical insulation purposes. (iii) PVC and calcium carbonate combination is used for general purposes such as tubings, seat covers, wires and cable insulations. (iv) Fibrous fillers such as wood flour, short length of synthetic fibres, cotton flock, macerated paper and cloth etc., are used in thermosetting polymers, phenol-formaldehyde, melamine, formaldehyde, melamine urea, etc., to get filled plastics of higher impact strength. (v) Asbestos-filled plastics are employed in electrical appliances. (vi) Fibre-reinforced plastics possess good thermal and shock resistance, good dimensional stability, good process ability and moulded articles can be repaired easily.

2.9.3 Application of Filled Plastics (Reinforced Plastics) Filled or reinforced plastics find numerous applications. Some examples are as follows: (i) In automobiles for making door handles battery cases, engine cooling fans, etc. (using base polymers polyethylene, polypropylene and nylon-66). (ii) In defence for making nose cones, pistol grips and riffle bullets (using base polymers polystyrene and nylon-66). (iii) In textiles for making shuttle (nylon-66 as base polymer). (iv) In electrical/electronic industry for making exhaust fan, computer tape, insulators, wire and cable insulation, switch gear parts, spools, etc., (using polypropylene, PET, nylon and SAN as base polymers.)

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(v) In consumer goods like door handles, and window, hinge, chair shells, camera housing etc. (using polypropylene and ABS as base polymers). (vi) Miscellaneous like water meter housings, chemical pump housings, tubings, seat coverings etc., (using nylon-66, PVC and polypropylene as polymers).

2.10 BIOPOLYMERS Biopolymers are macromolecules that occur in nature from plants, trees, bacteria, algae or other sources that are long chains of molecules linked together through a chemical bond. They are often degradable through microbial processes such as composting. For example, cellulose, proteins, starch, collagen, casein, polyesters, etc. “Sustainable biopolymers” are sourced from sustainably grown and harvested cropland or forests, manufactured without hazardous inputs and impacts, healthy and safe for the environment during use and designed to be reutilised at the end of their intended use through recycling or composting. The potential for using these materials to make synthetic polymers was identified in the early 1900s, but they have only recently emerged as a viable material for large-scale commercial use.

2.10.1 Major Feed Stocks for Biopolymers While biopolymers can be made from an almost unlimited range of bio-based materials, most of the currently marketed biopolymers are made from starch. Corn is currently the primary feedstock, with potatoes and other starch crops also used in lower amounts.

2.10.2 Preparation Methods At present, either renewable or synthetic starting materials may be used to produce biodegradable polymers. Two main strategies may be followed in synthesising a polymer. One strategy is to build up the polymer structure from a monomer by a process of chemical polymerisation. The alternative is to take a naturally occurring polymer and chemically modifying it to give it the desired properties. A disadvantage of chemical modification is that the biodegradability of the polymer may be adversely affected. Therefore, it is often necessary to seek a compromise between the desired material properties and biodegradability.

2.10.3 Important Biodegradable Polymers Biodegradable polymers can be easily degradable by biological activities of decomposers. The natural raw materials are abundant, renewable and biodegradable, making them attractive feedstock fora new generation of environmentally friendly bioplastics. Even if only a small percentage of the biopolymers already being produced were used in the production of plastics, it would significantly decrease our dependence on non-renewable resources. (i) Lactic acidis now commercially produced on a large scale through the fermentation of sugar feedstock obtained from sugarcane, from the conversion of starch from corn, potato peels or other starch source. It can be polymerised to produce poly (lactic acid). Poly (lactic acid)has become a significant commercial polymer, useful for recyclable and biodegradable packaging, such as bottles, yogurt cups and candy wrappers. It has also been used for food

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(ii)

(iii)

(iv)

(v)

(vi) (vii)

(viii)

service ware, lawn and food waste bags, coatings for paper and cardboard and fibres for clothing, carpets, sheets and towels and wall coverings. In biomedical applications, it is used for prosthetic materials, drug encapsulation, biodegradable medical devices and materials for drug delivery. Triglycerides is another promising raw material for producing plastics from a large part of the storage lipids in animal and plant cells such as soybean, flax and rape seed. Triglycerideshave recently become the basis for a new family of composites. With glass fibre reinforcement, they can be made into long-lasting durable materials with applications in the manufacture of agricultural equipment, the automotive industry, construction and other areas. Fibres other than glass can also be used in the process, such as fibres from jute, hemp, flax, wood and even straw or hay. If straw could replace wood in composites now used in the construction industry, it would provide a new use for an abundant, rapidly renewable agricultural commodity and at the same time conserve less rapidly renewable wood fibre. Starchis found in corn (maize), potatoes, wheat, tapioca (cassava) and some other plants. Starchbased bioplastics are important not only because starch is the least expensive biopolymer but because it can be processed by all of the methods used for synthetic polymers, like film extrusion and injection moulding. Utensils, plates, cups and other products have been made with starchbased plastics. The annual world production of starch is well over £70 billion, with much of it being used for non-food purposes, such as making paper, cardboard, textile sizing and adhesives. Starch-protein compositionshave the interesting characteristic of meeting nutritional requirements for farm animals. Hog feed, for example, is recommended to contain 13%–24% protein, and complemented with starch. If starch-protein plastics were commercialised, used food containers and service ware collected from fast food restaurants could be pasteurised and turned into animal feed. The interest insoybeanshas been revived, recalling Ford’s early efforts. In research laboratories, it has been shown that soy protein, with and without cellulose extenders, can be processed with modern extrusion and injection moulding methods and also used for making adhesives and coatings for paper and cardboard. Manywater-soluble biopolymerssuch as starch, gelatine, soy protein and casein form flexible films when properly plasticised. Although such films are regarded mainly as food coatings, it is recognised that they have potential use as non-supported stand-alone sheeting for food packaging and other purposes. Casein, commercially produced mainly from cow’s skimmed milk, is used in adhesives, binders, protective coatings and other products. Polyestersare now produced from natural resources such as starch and sugars, through large-scale fermentation processes, and are used to manufacture water-resistant bottles, utensils and other products. Polyestersare produced by bacteria, and can be made commercially on a large scale through fermentation processes. They are now being used in biomedical applications. Collagenis the most abundant protein found in mammals. Gelatine is denatured collagen, and is used in sausage casings, capsules for drugs and vitamin preparations and other miscellaneous industrial applications including photography.

2.10.4 Importance of Biopolymers in Sustainable Development There are enormous societal benefits that result from a shift to bio-based plastics. Bio-based materials have the potential to produce fewer greenhouse gases, require less energy and produce fewer toxic pollutants over their lifecycle than products made from fossil fuels. They may also be recyclable or

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composted (depending on the biomaterial and how it is produced), reducing waste streams to already crowded landfills or to incinerators. As the cost of petroleum increases, making products with bio-based materials is increasingly attractive. Increased demand for agricultural and forest-based feedstock also offers new resourcebased economic development opportunities for farmers, struggling rural communities and manufacturing sectors. However, many of these advantages are not inherent in the material. They depend on ensuring that bio-based products meet minimal standards for the safe production, use and end-of-life disposition. Making the transition from a petroleum-based to a bio-based economy also gives us an opportunity to ensure that the impact of product standards on the environment, health and society are included. The widespread use of these new plastics will depend on developing technologies that can be successful in the marketplace. This, in turn, will partly depend on how strongly society is committed to the concepts of resource conservation, environmental preservation and sustainable technologies. There are growing signs that people indeed want to live in greater harmony with nature and leave a healthy planet to the future generation. If so, bioplastics will find a place in the current age of plastics.

2.11 CONDUCTING POLYMERS Due to non-availability of free electrons, most normal polymers are insulators. Scientists have taken this property as an advantage, and with their curiosity and challenging nature, they prepared conducting polymers as promising materials. “Conducting polymer is an organic polymer having highly delocalised π-electron system and electrical conductance”. Conducting polymers are broadly classified into two categories as intrinsically conducting polymers and extrinsically conducting polymers.

2.11.1 Intrinsically Conducting Polymer (ICP) or Conjugated π-Electrons Conducting Polymer Polymers which contain conjugated π-electron backbone or delocalised electron pairs act as intrinsic conducting polymers. For example, (i) Polyacetylene polymers like Poly-p-phenylene, polyquinolone, etc. (ii) Condensed aromatic polymers like polyaniline, polyphenanthrylene, etc. (iii) Heteroaromatic and conjugated aliphatic polymers like polypyrrole, polyazomethine, etc. If the conductivity of intrinsically conducting polymers is less, they are doped with positive or negative charges and this process is known as doping. It is mainly oxidative or p-doping, reductive or n-doping and protonic acid doping. (i) Oxidative or p-doping: Treating of intrinsically conducting polymers with Lewis acid like I2, Br2, FeCl3, CCl4, HBF4, etc., is known as p-doping. (C2 H 2 ) n + 2FeCl3 → (C2 H 2 )+n FeCl 4− + FeCl 2 2(C2 H 2 ) n + 3I 2 → 2[(C2 H 2 )+n I3 ]

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2.38 Engineering Chemistry Mechanism of conduction: The removal of an electron from the polymer ∏-back bone using a suitable oxidising agent leads to the formation of delocalised radical ion called polarion. A second oxidation of a chain containing polarion, followed by radical recommendation, yields two charge carriers on each chain. The positive charges sites on the polymer chains are compensate by anions formed by the oxidising agent. The delocalised positive charges on the polymer chain are mobile, not the dopant anions. Thus, these delocalised positive charges are current carriers for conduction. These charges must move from chain to chain as well as along the chain for bulk conduction. (ii) Reductive (or) n-doping: Treating of intrinsically conducting polymers with Lewis bases such as Li, Na, tetrabutyl ammonia, naphthylamine, etc., is known as n-doping. CH

+ C10H7NH2

Polyacetylene

Reduction

Napthylamine (Lewis base)

−

+ C10H8 Naphthalene

Mechanism of conduction: The addition of an electron to the polymer ∏-back bone by using a reducing agent generates a polarion. A second reduction of chain containing polarion, followed by the recombination of radicals, yields two negative (-ve) carriers on each chain. These charge sites on the polymer chains are compensated by cations formed by the reducing agent. (iii) Protonic acid doping: The synthesis of conducting polyaniline is a typical example of this type of doping technique. In this technique, current-carrying charged species (-ve/+ve) are created by the protonation of imine nitrogen. Polyaniline is partially oxidised first, with a suitable oxidising agent, into a base form of aniline, which contains alternating reduced and oxidised forms of aniline polymer backbone. This base form of aniline when treated with aqueous HCl (IM) undergoes protonation of imine nitrogen atom, creating current due to +ve sites in the polymer backbone. These charges are compensated by the anions (Cl−) of the doping agent, giving the corresponding salt. This doping results increase conductivity up to 9–10 orders of magnitude. Applications Conducting polymers are the most important materials to be used in electric and electronic applications. Some of the uses are as follows: (i) As electrode material for commercial rechargeable batteries, for higher power-to-weight ratio. (ii) As conductive tracks on printed circuit boards. (iii) As sensors—gas sensor, radiation sensor, humidity sensor, bio sensor for glucose and galactose, etc. (iv) As film membranes for gas separations. (v) As light-emitting diode. (vi) In electrochemical display windows. (vii) In fuel cells, as the electro catalytic materials. (viii) In information storage devices.

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2.11.2 Conducting Polyaniline In 1985, Alan MacDiarmid investigated polyaniline as an electrically conducting polymer. N

Properties The properties of polyaniline are as follows: (i) Except specific conductivity, all other properties of polyaniline are considered as that of an organic metal. (ii) It is transparent in thin layers. (iii) When heated, it is stable in air. (iv) It is a highly reactive, redox active material. (v) In conducting state, it is green and changes its colour in different media. For example, under reducing condition, it becomes yellow and under oxidising or basic condition, it becomes blue. Advantages The advantages are as follows: (i) It is the foremost air-stable conducting polymer. (ii) It has a wide and controllable range of conductivity. (iii) It shows a number of interesting properties such as multicolour, electrochromism, chemical sensitivity and so on. (iv) It is considered to be unique among all the conducting polymers as it can be synthesised chemically or electrochemically as a bulk powder or film. Disadvantages The disadvantages are as follows: (i) It is insoluble in common solvents except strong acids and n-methyl pyrrolidine. (ii) It has poor mechanical properties. (iii) It decomposes prior to melting, and hence it is difficult to process. Uses The uses of intrinsically conducting polymers are as follows: (i) Due to its reversible electrochemical response during anodic oxidation and cathodic reduction, it is useful as a secondary electrode in rechargeable batteries and electrochromic display devices. (ii) It is used for coating of films and semi-finished articles. (iii) It is also used for corrosion protection, sensors, smart windows, printed circuit boards, conductive pipes for explosives, and conductive fabrics.

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2.11.3 Extrinsically Conducting Polymers Polymers whose conductivity is due to externally added ingredient are known as extrinsically conducting polymers. They are conductive element filled polymers and blended conducting polymers. (i) Conductive element-filled polymers: In this kind of polymers, conducting materials like carbon black, metal oxides, etc., are added to the polymers. (ii) Blended conducting polymers: In this kind of polymer, conducting polymers are mixed with conventional polymer. Application of Conducting Polymers Conducting polymers have many uses because they are light weight, easy to process and have good mechanical properties. They are used in the following: (i) In chargeable light-weight batteries based on perchlorate-doped polyacetylene lithium systems. These are approximately 10 times lighter than conventional, lead storage batteries. They have flexibility to fit into different designs. (ii) In optical display devices, based on polythiophene and polyaniline. When the structure is electrically biased, the optical density of the film and its colour changes. Such electrochromic systems produce colour displays with faster switching time and better viewing than conventional liquid display devices. Due to redox activity in conducting state, polyaniline is green; under reducing condition, it is in yellow colour, whereas under oxidising condition, it appears blue. (iii) In telecommunication systems (iv) In electromagnetic screening materials (v) In antistatic coatings for clothing (vi) In electronic devices such as transistors and diodes (vii) In wiring in aircrafts and aerospace components (viii) In solar cells and as biosensors in metabolic reactions and drug delivery system for the human body (ix) In photo voltaic devices (x) In non-linear optical material (xi) In molecular wires and molecular switches

2.12 Polyphosphazenes/Phosphonitrilic Polymers Polyphosphazenes are hybrid inorganic organic polymers with a number of different skeletal structures that contain a backbone of alternating phosphorous and nitrogen atoms, and are interesting, commercially promising materials. A variety of substituents can substitute the basic backbone and hence we can get a variety of products. The basic backbone of polyphosphazene is as follows: Cl N

P Cl

Preparation: The most popularly used method for preparing polyphosphazenes is ring opening and substitution method. Allcock and co-workers discovered that cyclic trimer (hexachlorocyclo triphosphazene) can be thermally ring opened and can give high molecular weight soluble poly (dichlorophosphazene). After the replacement of the chlorine atoms in poly (dichlorophosphazene) by reaction with organic/organometallic nucleophiles, they give a variety of polyphosphazenes.

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Polymers Cl

2.41

Cl P

Heat

Cl Cl

250°C

P Cl

Hexachlorocyclo triphosphazene

Poly (dichlorophosphazene)

Substitution Reactions (i) Reaction with sodium alkoxide: Poly (dichlorophosphazene) reacts with sodium alkoxide and gives poly (dialkoxyphosphazene). Cl N

P Cl

OR RONa

–NaCl

Poly (dialkoxyphosphazene)

(ii) Reaction with amine: Poly (dichlorophosphazene) reacts with amine and gives poly (dialkylaminephasphazene). Cl N

P Cl

NHR

RNH2

–HCl

NHR

Poly (dialkylaminephasphazene)

(iii) Reaction with metal alkyd: Poly (dichlorophosphazene) reacts with metal alkyd and gives poly (dialkylphasphazene). Cl R N

P Cl

RM –MCl

P R

Poly (dialkylphasphazene)

Important Characteristics of Polyphosphazenes polyphosphazenes have many important properties; biocompatibility, high dipole moment, flexibility, chemical inertness, broad range of glass transition temperature (Tg), elastomeric property and impermeability are the most important of them. Applications Based on their wide range of unique properties, polyphosphazenes have countless and advanced applications. They have potential for formation of new compounds. The applications include in challenging areas of biomedical research such as tissue generation, macromolecules and so on. These are also used as ion-conductive membranes for rechargeable lithium batteries and fuel cell membranes. These are

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2.42 Engineering Chemistry advanced materials of elastomers for aerospace engineering. Polyphosphazenes are good photonic materials and fire-resistant polymers.

2.13 COMPOSITES Composites are multiphase materials that exhibit a significant proportion of the properties of both the constituent materials. (or) Composite materials composed of at least two distinctly dissimilar materials act in harmony. A judicious combination of two or more distinct materials can provide better combination of properties or an artificially prepared multiphase material in which the chemically dissimilar phases are separated by a distinct interface. For example, wood is the composite of cellulose and lignin, bone is the composite of a soft, strong protein collagen, and brittle, hard apatite material. Packing paper impregnated with bitumen or wax, rain-proof cloth (cloth impregnated with waterproof material), insulating tape, reinforced concrete, etc.

2.13.1 Constituents of Composites Composite material mainly comprises of the following: (i) Matrix phase: Matrix phase is the continuous body constituent enclosing the composite and given in its bulk form. Depending upon the matrix phase, composites are known as ceramic matrix composites, metal matrix composites, polymer matrix composites, etc. Functions of Matrix Phase

(a) Matrix phase binds the dispersion phase, act as a medium, applied stress is transmitted and distributed uniformly. (b) It protects the surface from damage due to chemical reactions, mechanical abrasion, etc. (c) It prevents the propagation of brittleness, cracking, etc. Hence, a good matrix phase should be ductile, having corrosion resistant and possess high binding strength. (ii) Dispersed phase: Dispersed phase is the structural constituent of composite. Fibres, flakes, whiskers, etc., are some important dispersed phases.

2.13.2 Classification of Composites Composites are broadly classified into three categories: (i) Particle-reinforced composites: In these composites, the dispersed phase is equiaxed, that is, the dimensions of the particles are nearly the same in all directions. They are subdivided into the following: (a) Large-particle composites (b) Dispersion-strengthened composites (ii) Fibre-reinforced composites: In these composites, the dispersed phase is in the form of fibres. These are subdivided into (a) continuous aligned (b) discontinuous. Discontinuous composites are further divided into (a) aligned (b) randomly oriented.

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(iii) Structural composites: In these composites, the properties depend not only on the constituent material but also on this geometrical design. These are subdivided into (a) laminates and (b) sandwich panels. Among these, fibre-reinforced polymer composites are widely used. Fibre-reinforced Polymer Composites These are prepared by reinforcing a plastic matrix with a high-strength fibre material. Fibre-reinforced composites involve three components, namely filament, a polymer matrix and an encapsulating agent (which ties fibre filaments to polymer). Glass fibres and metallic fibres are commonly employed for this purpose. The fibres can be employed either in the form of continuous lengths, staples or whiskers. Such composites possess high specific strength (tensile strength/specific gravity) and high specific modulus (elastic modulus/specific gravity), stiffness and lower overall density. Characteristics The fibre-reinforced composites possess superior properties such as higher yield strength, facture strength and fatigue life. The fibres prevent slip and crack propagation and inhibit it, thereby increasing mechanical properties. When a load is applied, there is a localised plastic flow in the matrix, which transfers the load to the fibres embedded in it. When a soft phase is present in hard matrix, the shock resistance of the composite is increased. On the other hand, if hard-reinforcing fibres are present in a soft matrix, the strength and modulus of the composite are increased. To obtain composites having the maximum strength and modulus, it is essential that there should be maximum number of fibres per unit volume, so that each fibre takes its full share of the load. The fibre-reinforced composites are, generally, anisotropic (i.e., having different directions), and the maximum strength is in the direction of alignment of fibres. For getting isotropic properties, the fibres are oriented randomly within the matrix, for example, ordinary fibre glass. It may be pointed here that the cost of laying fibres aligned in a particular direction is much higher than that for random orientation. For preparing fibre-reinforced composites, the following are essential: (i) The coefficient of expansion of the fibre matches closely that of the matrix. (ii) The fibre and matrix should be chemically compatible with each other and no undesirable reaction takes place between them. (iii) The fibre should be stable at room temperature and should retain a good percentage of strength at elevated temperatures. Some important reinforced composites are described here. (i) Glass fibre-reinforced polymer composites: For improving the characteristics of nylon, polyester, etc., containing polymer matrices, glass fibres are employed. These have lower densities, higher tensile strengths and resistance to corrosion and chemicals. Limitations: The limitations are as follows: (a) Since the most polymeric matrices start deteriorating or flow or melt at high temperatures, they find application with limited temperature service conditions. (b) They cannot be employed as structural components, since these materials do not possess the desired stiffness and rigidity.

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2.44 Engineering Chemistry Applications: They are used in automobile parts, storage tanks, floorings (industrial), transportation industries, plastic, pipes, etc. (ii) Carbon fibre-reinforced polymer composites: They are also known as advance polymer matrix composites or high performance composites and are employed in situations requiring (i) excellent resistance to corrosion, (ii) lighter density and (iii) retention of desired properties, even at elevated temperatures. However, the general use is limited due to their higher costs. Applications: They are used as structural components (like wing, body and stabiliser) of aircrafts (military and commercial) and helicopter’s recreational equipment (fishing rod), sport materials (golf clubs), etc.

2.13.3 Advantages of Composites over Conventional Materials Composites have the following advantages over conventional materials such as metals, polymers, ceramics and so on. (i) They have higher specific strength and stiffness. They can maintain strength up to higher temperatures. (ii) They have lower specific gravity, electrical conductivity and thermal expansion. (iii) They have better toughness, impact, thermal shock resistance, fatigue strength, corrosion and oxidation resistance.

2.13.4 Applications of Composites (i) In automobile industries, transportation industries, turbine engines, wire drawing dies, valves, pump parts, spray nozzles, storage tanks, fabrication of roof and floors, furniture, sport goods (lawn, tennis rackets), high-speed machinery, etc. (ii) Marine applications like propellers, shafts, hulls, spars (for racing boats) and other ship parts. (iii) Aeronautical applications like components of rockets, aircrafts (business and military), helicopters, missiles, etc. (iv) Communication antennae and electronic circuit boards (v) Safety equipment like ballistic protection

2.14 REVIEW QUESTIONS 2.14.1 Fill in the Blanks 1. Repeating units present in polymer is _______. [Ans.: monomer] 2. When two or more different monomer units participate in polymerisation, _______ polymers are formed. [Ans.: copolymers] 3. _______ plastics cannot be remoulded. [Ans.: thermosetting]

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4. Monomer units that participate in the formation of Fluon/Teflon are called _______. [Ans.: tetrafluoroethylene] 5. Bakelite is _______ plastic. [Ans.: thermosetting] 6. _______ is the basic unit present natural rubber [Ans.: Isoprene] 7. _______ monomers are involved in formation of nylon 66. [Ans.: Hexamethylenediamine and Adipic acid] 8. Monomers possess the double bonds undergo _______ polymerisation mechanism. [Ans.: addition] 9. An organic polymer with a highly delocalised pi-electron system, having electrical conductance is called a _______. [Ans.: conducting polymer] 10. _______ is an organic conducting polymer [Ans.: Polyaniline] 11. The polymers which occur in nature from plants, bacteria and algae are called _______. [Ans.: Bio polymer] 12. Triglycerides are produced from _______. [Ans.: soya bean]

2.14.2 Multiple-choice Questions 1. High polymers are (a) Liquids (c) Solids [Ans.: d]

(b) Gases (d) Colloidal

2. Polymerisation in which two or more chemically different monomers takes part, is called (a) Addition polymerisation (b) Co polymerisation (c) Chain polymerisation (d) None of these [Ans.: b] 3. Structural units of high polymers, are called (a) Fibres (b) Thermo units (c) Monomers fabrics (d) Fabrics [Ans.: c] 4. GR-S rubber is an example of (a) Condensation polymerisation (c) Addition polymerisation [Ans.: b]

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2.46 Engineering Chemistry 5. A thermoplastic is formed by the phenomenon of (a) Chlorination (b) Condensation polymerisation (c) Nitration (d) Chain polymerisation [Ans.: d] 6. A copolymer is formed by the union of two or more (a) A molecules of some monomer by the elimination of small molecules like H2O (b) Monomer molecules in a chain without the elimination of H2O etc. (c) Monomer molecules with the elimination of small molecules like H2O (d) Thermosetting and thermoplastic resins [Ans.: b] 7. Phenol-formaldehyde resin is commercially known as (a) PVC (b) Elastomer (c) Bakelite (d) Nylon [Ans.: c] 8. Polymer commonly used for making fibre/cloth is (a) Rubber (b) PVC (c) Nylon (d) Bakelite [Ans.: c] 9. Bakelite is prepared by the condensation of (a) Benzene and formaldehyde (b) Phenol and formaldehyde (c) Phenol and acetaldehyde (d) Glycerol and phthalic acid [Ans.: b] 10. A high molecular weight material that can easily be moulded in any desired shape is (a) Graphite (b) Resin (c) Jelly (d) Grease [Ans.: b] 11. A plastic which can be softened on heating and hardened on cooling is called (a) Thermoelastic (b) Thermoplastic (c) Thermosetting (d) Thermite [Ans.: b] 12. Phenol-formaldehyde (Bakelite) is an example of (a) Thermoelastic (b) Thermoplastic (c) Thermosetting (d) Thermite [Ans.: c] 13. Natural rubber is basically a polymer of (a) Propylene (c) Ethylene [Ans.: b]

(b) Isoprene (d) Propane

14. Which one of the following is an elastomer? (a) PVC (b) Bakelite (c) Natural rubber (d) Nylon [Ans.: c]

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Polymers 15. Raw rubber on vulcanisation becomes (a) Plastic (c) Soft [Ans.: d]

2.47

(b) Rocky (d) Less elastic

16. The most commonly used vulcanising agent is (a) Graphic (b) Carbon black (c) Sulphur (d) Dry ice [Ans.: c] 17. Ebonite is (a) Polyethylene (c) Natural [Ans.: b]

(b) Highly vulcanised rubber (d) Synthetic rubber

18. One of the important uses of Bakelite is for making (a) Cables (b) Electrical switches (c) Cloth (d) Hose pipe [Ans.: b] 19. The fibre obtained by condensation of hexamethylenediamine and adipic acid is (a) Decorn (b) Nylon (c) Rayon (d) Terylene [Ans.: b] 20. Nylon is (a) A polythene derivative (c) A polyamide fibre [Ans.: c]

(b) A polyester fibre (d) None of these

21. Monomer units involved in the formation of Bakelite are (a) Urea and formaldehyde (b) Phenol and formaldehyde (c) Urea and phenol (d) Butadiene and ethylene [Ans.: b] 22. Buna-s is a co polymer of (a) Vinyl chloride and vinyl alcohol (c) Butadiene and styrene [Ans.: c]

(b) Butadiene and acrylonitrile (d) Butadiene and ethylene

23. Which one is not a macro molecule? (a) Protein (c) Ice [Ans.: c]

(b) Insulin (d) Cellulose

24. Which one is a co polymer? (a) Nylon-66 (c) PVC [Ans.: a]

(b) Teflon (d) Polybutadiene

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2.48 Engineering Chemistry 25. An example of chain-growth polymer is (a) Nylon-66 (c) Terylene [Ans.: d]

(b) Bakelite (d) Teflon

26. What is the repeating unit in Teflon? (a) F2C = CF2 (c) H2C = CHCl [Ans.: a]

(b) H2C = CH2 (d) H2C = CHC6H5

27. Which of the following is not a thermoplastic? (a) Bakelite (b) Terylene (c) Polystyrene (d) Polythene [Ans.: a] 28. Which one is a step-growth polymer? (a) Teflon (c) Polybutadiene [Ans.: d]

(b) PVC (d) Bakelite

29. Which of the following is a synthetic polymer? (a) Cellulose (b) PVC (c) Proteins (d) Nucleic acids [Ans.: b] 30. Which of the following is a linear polymer? (a) Nylons (b) Bakelite (c) Low density polythene (d) Formaldehyde polymer [Ans.: a] 31. Which of the following has the largest molecular mass? (a) Monomer (b) Dimer (c) Oligomer (d) Polymer [Ans.: d] 32. Isoprene is monomer of (a) Starch (c) Natural rubber [Ans.: c]

(b) Synthetic rubber (d) PVC

33. Which of the following is thermosetting polymer? (a) Linear (b) Branched chain (c) Cross-linked (d) Thermoplastic [Ans.: c] 34. Which of the following is an addition polymer? (a) Terylene (b) Nylon (c) Polyethylene (d) All of these [Ans.: c]

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35. Which of the following is thermosetting polymer? (a) Nylon (b) Polyethylene (c) Bakelite (d) PVC [Ans.: c] 36. The monomer of PVC is (a) Ethylene (c) Tetrafluoroethylene [Ans.: b]

(b) Vinyl chloride (d) Styrene

37. Chloroprene is the repeating unit in (a) Polystyrene (c) PVC [Ans.: b]

(b) Neoprene (d) Polythene

38. The process of vulcanisation makes rubber (a) Soluble in water (b) Hard (c) Soft (d) More elastic [Ans.: b] 39. A common catalyst used in addition polymerisation is (a) Nickel (b) Y-zeolite (c) Ziegler–Natta catalyst (d) Platinum [Ans.: c] 40. Functionality of phenol is (a) One (c) Three [Ans.: c]

(b) Two (d) Six

41. _______ polymer is formed from the fermentation of sugar feed stocks obtained from sugarcane. (a) Lactic acid (b) Steric acid (c) Triglycerides (d) None of these [Ans.: a] 42. Adhesive, binders and protective coating characteristics exhibiting polymers is (a) Lactose (b) Galactose (c) Starch (d) Casein [Ans.: d] 43. Water soluble bio-polymers is/are (a) Starch (c) Both (a) and (b) [Ans.: a]

(b) Polyester (d) None of these

2.14.3 Short Answer Questions 1. What is a monomer? Ans.: The repeating unit present in the formation of a polymer is known as a monomer.

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2.50 Engineering Chemistry 2. What is a polymer? Ans.: Polymers are macromolecules of high molecular masses built up by the linking together a large number of small, repeated units by a covalent bond. 3. What is meant by polymerisation? Ans.: The chemical process leading to the formation of a polymer is known as polymerisation. 4. What is meant by degree of polymerisation? Ans.: The number of monomer units in a polymer is known as the degree of polymerisation. 5. Differentiate between homopolymer and copolymer. Ans.: Homopolymers are formed with same monomer units. For example, PE, PS, PVC, etc. Copolymers are formed with two or more different monomers. For example, styrene butadiene rubber (styrene + butadiene) 6. Differentiate between addition and condensation polymerisation. Ans.: Addition polymerisation is the process of polymerisation by the addition of monomer units which have unsaturated double or triple bonds. For example, polyethylene, polyvinyl chloride, etc. Condensation polymerisation takes place where the monomer units have two or more reactive functional groups. For example, polyester, nylon, polyamide, etc. 7. Define thermoplastics. Ans.: Linear, long chain polymers which can be softened on heating and hardened on cooling are known as thermoplastics. For example, polythene, polyvinyl chloride, etc. 8. What are the monomers present on nylon 66? Ans.: Hexamethylenediamine and adipic acid. 9. What is meant by vulcanisation of rubber? Ans.: Vulcanisation is heating of the raw rubber at 100–140°C with sulphur. 10. What is natural rubber? Ans.: Cis-polyisoprene is a natural rubber. 11. What is gutta percha? Ans.: Trans-polyisoprene is gutta percha. 12. What is crepe rubber? Ans.: Rubber with sodium bisulphite is passed through a creping machine and the coagulum is rolled into sheets. The sheet is hence having the surface like crepe paper; hence, it is known as crepe rubber. 13. What is an elastomer? Ans.: A n elastomer is vulcanisable rubber like polymer, which can be stretched to at least twice its length and returned to its original shape and dimensions as soon as stretching force is released.

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14. Why does raw rubber need vulcanisation? Ans.: In vulcanisation, sulphur combines chemically at the double bonds of different rubber spring and provides cross-linking between the chains. Hence, for stiffening the rubber needs vulcanisation. 15. Give examples for natural polymers. Ans.: Cotton, silk, wool, nucleic acid, proteins, starch, cellulose, etc. 16. Give examples for inorganic polymers. Ans.: Polyphosphazenes, polysilanes, polygermanes, etc. 17. What are the substances mixed in the compounding of rubber? Ans.: A ntioxidants, colouring agents, vulcanising agents, accelerators, plasticisers and inert fillers are adding in the compounding of raw rubber. 18. Give note on Bio-polymers? Ans.: Bio-polymers are macromolecules that occur in nature from plants, tress, bacteria, algae or others source that are long chains linked together through a chemical bond. They are often degradable through microbial process such as compositing. For example, cellulose, proteins, starch, collagen, casein and polyesters 19. Write important bio-degradable polymers? Ans.: (a) Lactic acids: It is obtained from fermentation of sugar feed stocks conversion of starch from corn, potato peels, etc. (b) Triglycerides: It is produced from soya bean, fax and rape seed. (c) Starch: It is found in corn, potatoes, wheat, tapioca and some other plants. (d) Collagen: It is found in mammals. It is used in capsules for drug. 20. Explain the importance of bio polymers? Bio-based materials have the potential to produce fewer greenhouses gases, require less energy and produce fewer toxic pollutants.

(1) They may also be recyclable or composted. (2) They reduce waste stream.

2.14.4 Descriptive Questions Q.1 (a) Write a note on the properties and uses of Teflon. (b) Differentiate between natural polymer and synthetic polymer. (c) Write a note on silicone rubbers. Q.2 (a) Distinguish between addition and condensation polymerisation. (b) Explain the differences between thermoplastics and thermosetting plastics with examples. (c) What is meant by degree of polymerisation? Q.3 Write the structures of four addition polymers and four condensation polymers with their respective monomers. Q.4 (a) Describe the preparation, properties and engineering uses of polyethylene. (b) What is meant by fabrication of plastics? Mention the different fabrication techniques.

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2.52 Engineering Chemistry Q.5 (a) What are elastomers? Give examples. (b) What are the ingredients used in the compounding of plastics? What are their functions? Q.6 (a) What is a plastic? (b) Write the merits and demerits of using plastics in the place of metals. Q.7 (a) Identify the thermosets and thermoplastics among the following: (i) PVC (ii) Polyethylene (iii) Silicone (iv) Polyester fibre (v) Bakelite (b) What is Bakelite? How is it manufactured? Mention its uses. Q.8 (a) Explain the process of extrusion moulding with a neat diagram. (b) How are the following polymers prepared? Mention their properties and uses. (i) PVC (ii) LDPE Q.9 (a) What are elastomers? Give the preparation, properties and uses of Buna-S. (b) Describe a method for moulding of thermoplastic resin. Q.10 (a) Explain the preparation, properties and uses of Bakelite. (b) Describe the process of compression moulding with a neat sketch. Q.11 (a) W hy are silicones called inorganic polymers? Discuss the synthesis of linear chain silicones. (b) Why can Bakelite not be remoulded? Write its repeating unit. (c) Describe condensation polymerisation with an example. Q.12 (a) What is a homochain polymer? Give examples. (b) What is polymerisation? Explain the different types of polymerisation with examples. Q.13 (a) How is HDPE prepared? Give its properties and uses. (b) Explain the injection moulding process with a neat diagram. Mention its advantages. Q.14 Write informative notes on the following: (a) Classification of polymers (b) Mechanism of radical polymerisation (c) Anionic and cationic polymerisation (d) Thermodynamics of a polymerisation process Q.15 What are the common constituents of plastics and what are their functions? Q.16 Write short notes on the following: (a) Vulcanisation of rubber (b) Polyvinyl chloride (c) Compounding of rubber (d) Reclaimed rubber Q.17 Write an essay on the preparation, properties and uses of the following: (a) Polyvinyl acetate (b) Cellulose acetate

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(c) Phenol formaldehyde resins (d) Teflon (e) Polyethylene (f) Polystyrene (g) Polymethylmethacrylate Q.18 Discuss the various polymers related to natural rubber with emphasis on their preparation, properties and uses. Q.19 Write an account of application of polymers in bio-medical electronic fields. Q.20 Write an essay on the various types of synthetic rubber with brief description of the preparation, properties and uses of any three of them. Q.21 Write an essay on fibre-reinforced plastics. Q.22 What type of rubber would you recommend for the following? (a) For a flexible connection to a steam line (b) As a gasket for a pipe containing a chlorinated solvent (c) In a solvent to form an adhesive Q.23 Write short notes on the following: (a) Molecular weights of polymers (b) Conductive polymers (c) Ionic polymers (d) Liquid crystal polymers (e) Engineering polymers (f) Photo conductive polymers (g) Polymer structure and properties of polymers Q.24 Define the following terms and give examples: monomer, polymer, polymerisation and degree of polymerisation. Q.25 Distinguish clearly between the following: (a) Thermoplastic and thermosetting plastic (b) Natural and synthetic rubber (c) Addition and condensation polymerisation Q.26 Discuss the addition polymerisation reaction mechanism. Q.27 Explain condensation polymerisation with an example. Q.28 Write informative notes on the following: (a) Vulcanisation of rubber (b) Compounding of rubber (c) Gutta percha (d) Silicone rubber Q.29 Give the preparation, properties and engineering uses of the following: (a) Teflon (b) Polystyrene (c) Polyethene

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2.54 Engineering Chemistry (d) Polyurethane (e) HDPE (f) Conductive polymers Q.30 Discuss the various polymers related to natural rubber with emphasis on their preparation, properties and uses. Q.31 Write an essay on the various types of synthetic rubbers. Q.32 What are elastomers? Write the structure for natural rubber and gutta percha. Q.33 What is vulcanisation of rubber? Mention its uses. Explain why natural rubber needs vulcanisation. How is it carried out? Q.34 How is crepe rubber obtained from latex? Q.35 What is latex? How is natural rubber isolated from it? Q.36 Write short notes on silicones. How are they prepared? Q.37 Write a short note on conducting polymers. Q.38 Give the structural unit of gutta percha. Q.39 Give detailed notes bio-polymers and its importance. Q.40 Explain the preparation methods of bio-polymer and the importance of bio-degradable polymers.

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Fuels and Combustion “A revolution in humanity’s use of fossil fuel-based energy would be necessary sooner or later to sustain and to extend modern standards of living.”

3.1 INTRODUCTION ‘Fuel is the source of heat energy, it can be stored as potential chemical energy and can be released through combustion.’ ‘Combustible matter having carbon as a major ingredient, produce large amount of heat energy on burning and can be used for heat generation in industry and domestic applications is known as a fuel.’ ‘Any compound or a substance which can produce energy and can be used in the production of power is termed as a fuel.’ In the combustion process, a fuel reacts with oxygen and releases the energy. Fuel + O2 → CO2 + H2O + heat

3.2 CLASSIFICATION OF FUELS Fuels are broadly classified according to their occurrence and state of aggregation. According to the occurrence they are classified as primary (natural) and secondary (derived) fuels and based on the state of aggregation solid, liquid and gaseous fuels. Fuel

(Occurance)

Secondary (or) der ive State of aggregation

Primary (or) natural State of aggregation

Solid Wood, Peat Coal, Lignite

Liquid Petroleum (or) Crude oil

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Gaseous Natural gases

Solid Coke, semi coke, charcoal, petroleum coke, thiokol

Liquid Gasoline, diesel oil, Kerosene

Gaseous Coal gas water gas producer gas, LPG

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3.2 Engineering Chemistry

3.3 UNITS OF HEAT (i) CGS system–calorie: The amount of heat required to raise the temperature of 1 g of water through 1°C at 15 to 16°C. (ii) MKS system–kilo calorie: The amount of heat required to raise the temperature of 1 kg of water through 1°C. 1 kcal = 1,000 cal (iii) British system (FPS system): British thermal unit (BTU): The amount of heat required to raise the temperature of 1 pound of water through 1°F at 60 to 61°F. 1 BTU = 252 cal = 0.252 kcal 1kcal = 3.968 BTU (iv) Centigrade heat unit (CHU): The amount of heat required to raise the temperature of 1 pound of water through 1°C. 1 kcal = 3.968 BTU = 2.2 CHU

3.4 Calorific Value The total quantity of heat liberated by the complete combustion of one unit mass/volume of fuel in oxygen is known as calorific value. This is mainly divided into higher calorific value and lower calorific value. (i) Higher (or) gross calorific value (HCV or GCV): The higher or gross calorific value is the amount of heat liberated when one unit mass/volume of the fuel is burnt completely and the combustible products are cooled to room temperature. i.e., 25°C or 77°F. (ii) Lower or net calorific value (LCV or NCV): Lower calorific value is defined as the amount of heat liberated when one unit of fuel is burnt completely but the combustible products are allowed to escape; hence, here lesser amount of heat is available. LCV = HCV - latent heat of water vapour LCV = HCV - mass of hydrogen × 9 × latent heat of steam (587 kcal/kg) This is because one part by mass of hydrogen gives nine parts by mass of water. (iii) Units of calorific value:

LCV = HCV - 0.09HX 587 kcal/kg C.G.S-Calories/gram

For Solid and Liquid Fuels

M.K.S-Kcal/kg F.P.S-B.T.U/lb C.G.S-Calorie/cubic centimeter (cal/cm3)

Gaseous fuels

M.K.S-Kcal/cubic meter (kcal/m3) F.P.S-B.T.U/cubic feet (B.T.U/ft3)

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Fuels and Combustion

3.3

3.5 Determination of calorific value A calorimeter is used for determining calorific value. For determining calorific value of solid and liquid fuels a bomb calorimeter is used and for gaseous fuel Junker’s calorimeter is used.

3.5.1 Bomb Calorimeter Description Bomb calorimeter consists of strong cylindrical stainless steel bomb with lid. The bomb carries the fuel, and the lid can be screwed to the body of the bomb and make a perfect gas tight seal. The lid has two stainless steel electrodes and an oxygen inlet valve, and among these a small ring is attached to one of the electrodes. A nickel or stainless steel crucible is supported by that right. The bomb is placed in a copper colorimeter, for preventing heat loss by radiation, it is surrounded by air and water jacket. Stirrer which can operated electrically and Beckmann’s thermometer, having sensitivity to read up to 0.01°C are provided. The set-up is shown in Figure 3.1. 6V battery Beckmann’s thermometer

Oxygen valve Electrically operated stirrer

Electrodes to which a ring is attached

Copper calorimeter

Mg fuse wire

Stainless steel bomb

Weighted pallet of given fuel sample

Stainless steel crucible Air jacket Water jacket

Figure 3.1 Bomb calorimeter Working In the clean crucible, a weighted amount (0.5 to 1.0 g) of the fuel is taken and the crucible is supported by a ring, a fine magnesium wire touching the fuel sample is stretched across the electrodes. The bomb lid is tightly screwed, filled with oxygen to 25 atm pressure and then lowered into copper calorimeter, containing known mass of water, and the initial temperature (t1) is noted. Now, the circuit is completed by connecting the electrodes with a 6 V battery. The sample burns, liberates heat and absorbed by water. The water is stirred continuously for maintaining uniform temperature, and hence the final temperature (t2) is noted. Observations and Calculation Weight of fuel = x g Weight of water in calorimeter = w g

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3.4 Engineering Chemistry Water equivalent of calorimeter = wt. of apparatus × specific heat = w g Initial temperature of water = t °C Final temperature of water = t2 °C High calorific value of the fuel (w + w ) (t2 − t1 ) HCV = cal/g x Lower calorific value of the fuel LCV = HCV – 0.09 H × 587 cal/g Latent of heat of steam = 587 cal/g Weight of water produced from 1 g of fuel = 9H/100 g = 0.09H g H = percentage of hydrogen in fuel.

3.5.2 Junker’s Calorimeter Junker’s gas calorimeter (Figure 3.2) consists of a vertical cylindrical combustion chamber, and the pressure governor regulates the supply of gaseous fuel. Gasometer measures the volume of gas flowing in a particular time and combustion of fuel can be carried out by a Bunsen’s burner. The combustion chamber is surrounded by an annular water space, inside heat exchange coils and outer flues are fitted. Chromium plated outer jacket which prevent the radiative and convective heat loss from calorimeters because it contains air and acts as a very good insulator. Openings of annular space can circulate the water at the appropriate places at constant rate around the combustion chamber. Two thermometers placed at appropriate place can measure the temperatures of the inlet and outlet water. Thermometer (T2)

To measuring vessel

Cylindrical combustion chamber

ome t The rm

Thermometer (T1)

Outer jacket Air

Water from constant head

Exit gases Condensate Gas

Bunsen burner Gasometer

Figure 3.2 Junker’s gas calorimeter

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Fuels and Combustion

3.5

In the combustion chamber a known volume of gas is burned at a constant rate in excess of air, produced heat is absorbed by water. From the temperature difference, heat evolved from the gas can be calculated. Observations and Calculation Volume of the gas burnt in ‘t’ at STP = V Temperature of incoming water = T1 Temperature of outgoing water = T2 Weight of water collected in time t = w w(T2 − T1 ) High calorific value = kcal/m3 . V Mass of steam condensed in time ‘t’ in graduated cylinder from V m3 of gas = m Latent heat of steam = 587 cal/kg m   Lower calorific value = LCV =  HCV − × 587 kcal/m3 v   LCV = HCV − 0.09XHX587 kcal/kg Calculation of calorific value of a fuel can be made theoretically by using Dulongs formula. Solved Numerical Problems Based on Calorific Value (i) Calculate the gross and net calorific value of coal having the following compositions carbon – 85%, hydrogen – 8%, sulphur – 1%, nitrogen – 2%, ash – 4%, latent heat of steam – 587 ca/g. Solution Gross Calorific Value (GCV) =

1  0   8080 × C + 34, 500  H −  + 2, 240 × S k cal/kg    100  8 

0 1    8080 × 85 + 34, 500  8 −  + 2, 240 × 1 k cal/kg  100  8  1 [686, 800 + 276, 000 + 2, 240] k cal/kg = 100 1 = [965040] k cal/kg = 9650.4 k cal/kg 100 =

Net Calorific Value (NCV) = (GCV – 0.09H × 587) k cal/kg = (9650.4 – 0.09 × 8 × 587) k cal/kg = 9227.8 k cal.kg (ii) A coal has the following composition by weight: C – 90%, O – 3.0%, S – 0.5%, N = 0.5% and ash = 2.5%. Net calorific value of the coal was found to be 8490.5 k cal/kg. Calculate the percentage of hydrogen and higher calorific value of coal. Solution HCV = (HCV + 0.09H × 587) k cal/kg = (8490.5 + 0.09H × 587) k cal/kg = (8490.5 + 52.8H) k cal/kg (i) 1  3.0    8080 × 90 + 34, 500  H −  + 2, 240 × 0.5 k cal/kg   100  8  = [7272 + 345H − 129.4 + 11.2] k cal/kg = [7754.8 + 345H] k cal/kg (ii)

Also HCV =

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3.6 Engineering Chemistry From (i) and (ii), we get

7754.8 + 345H = 8490.5 + 52.8 H or 292.2H = 8490.5 – 7154.8 = 1335.7 or percentage of H = 1335.7/292.2 = 4.575% HCV = (8490.5 + 52.8 × 4.575) k cal/kg [From (i) and (iii)] = (8490.5 + 241.3) k cal/kg = 8731.8 k cal/kg

(iii)

(iii) 0.72 gram of a fuel containing 80% carbon, when burnt in a bomb calorimeter, increased the temperature of water from 27.3° to 29.1°C. If the calorimeter contains 250 gm of water and its water equivalent is 150 gm, calculate the HCV of the fuel. Give your answer in kJ/kg. Solution Here x = 0.72 gm, W = 250 gm, w = 150 gm, t1 = 273°C, t2 = 29.1°C (W + w )(t 2 − t1 ) k cal/kg x (250 + 150) × (29.1− − 27.3) = k cal/kg 0.72 = 1000 × 4.2 kJ/kg = 4200 kJ/kg

∴ HCV of fuel (L) =

(iv) On burning 0.83 g of a solid fuel in a bomb calorimeter, the temperature of 3500 g of water increased from 26.5°C to 29.2°C. Water equivalent of calorimeter and latent heat of steam are 385.0 g and 587.0 cal/g respectively. If the fuel contains 0.7% hydrogen, calculate its gross and net calorific value. Solution Here, wt. of fuel (x) = 0.83 g; wt of water (W) = 3500 g; water equivalent of calorimeter (w) = 385 g; (t2 – t1) = 2.7°C; percentage of hydrogen (H) = 0.7%; latent heat of steam = 587 cal/g. ( W + w )( t 2 − t1 ) (3500 + 385) × 2.7 \ Gross calorific value = = = 12638 cal/g 0.83 x Net calorific value = [GCV – 0.09H × 587] = (12683 – 0.09 × 0.7 × 587) cal/g = (12683 – 37) cal/gm = 12601 cal/gm (v) Calculate the calorific value of a fuel sample of the coal with the following data. Mass of the coal = 0.6 g Water equivalent of calorimeter = 2200 gm Specific heat of water = 4.187 kJ kg−1 °C−1 Rise in temperature = 6.52 °C Solution Heat liberated by burning 0.6 g coal = 3.3 kg × 4.187 kJ kg−1 °C−1 × 6.52 °C = 60.06 kJ \ Calorific value of coal = 60.06 kJ/0.6 g = 100.1 kJ g−1. (vi) A sample of coal contains C = 93% H = 6 % and ash = 1%. The following data were obtained when the above coal was tested in bomb calorimeter. (a) Wt. of coal burnt – 0.92 gm (b) Wt. of water taken – 550 gm

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3.7

(c) Water equivalent of bomb & calorimeter – 2200 g (d) Rise in temperature – 2.42 °C (e) Fuse wire correction – 10.0 cal (f) Acid correction – 50.0 cal Calculate grass and net calorific values of the coal, assuming the latent heat of condensation of steam as 580 cal/g. Solution Wt. of coal sample (x) = 0.92 g: wt. of water (W) = 550 g: water equivalent of calorimeter (w) = 2200 g: temperature rise (t2 – t1) = 2.42 °C; acid correction = 50.0 cal, fuse wire correction = 10.0 cal; latent heat of steam = 580 cal/g percentage of H = 6% ( W + w )(t2 − t1 ) − [Acid + fuse wire correction ] x (550 + 2200)(2.42) − (50 + 10) cal/g = 0.92 = 7168.5 cal/g

∴ GCV =

NCV = (GCV − 0.09H × Latent heat of steam ) − (7168.5 − 0.09 × 6 × 580) cal/g = 6855.3 cal/g

3.6 Characteristics of Good Fuel Important characteristics of a good fuel are listed hereunder. (i) HCV: The amount of heat released is dependent on high calorific value, hence fuel should possess more HCV. (ii) Low moisture content: Moisture content of fuel reduces the calorific value, hence fuel should possess low moisture content. (iii) Moderate ignition temperature: Minimum required temperature to preheat the fuel and starts burning is the ignition temperature. Fuel must have moderate ignition temperature, because low ignition temperature is dangerous for storage and transport due to fire hazard and for starting a fire, high ignition temperature is not suitable. (iv) Moderate velocity of combustion: For continuous supply of heat, fuel must burn with a moderate velocity. (v) Low non-combustion matter and ash content: After combustion, non-combustible matter produces high ash content and also reduces the heating value. With this more heat loss, and loss of money for over storage, handling, disposal of ash, etc. (vi) Low cost: Good fuel should be available easily in bulk at low cost. (vii) High pyrometric effect: Pyrometric effect is the highest temperature obtained with the fuel, hence fuel should have high pyrometric effect. (viii) Less pollutants/environmental friendly: By-products of combustion like CO, SO2, NO2, etc. pollute the environment, so a good fuel should release less pollutants and should be environment friendly. (ix) Storage cost: Storage cost of a fuel in bulk should be low (x) Easy transportation: Fuel should be easy to transport with low cost. (xi) Uniform particle size: In case of solid fuels, for easy combustion, the particle size should be uniform.

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3.8 Engineering Chemistry

3.7 Solid Fuels Coal and coke are main solid fuels.

3.7.1 Coal Coal is the primary and largest solid fuel used to produce electricity and heat through combustion. Black or brownish black sedimentary rock usually occur as coal beds, composed primarily of carbon along with other elements like hydrogen, oxygen, nitrogen and sulphur, also known as pulverised carbon. Due to biogeological processes, from the dead plant matter and vegetation fossil fuel coal is formed, and is slowly converted into peat, lignite, bituminous coal and finally to anthracite. According to carbon and hydrogen ratio, ranking of coal in increasing order is as follows Peat  → Lignite  → Semibitumiuous coal  → Bituminous coal  → Anthracite  → Graphite

3.7.2 Analysis of Coal Proximate and ultimate analysis is carried out to assess and determine the quality of coal. 3.7.2.1 Proximate Analysis of Coal Practical utility of coal is determined by the proximate analysis. Here, information is obtained regarding moisture, volatile matter, ash and fixed carbon content. (i) Moisture: In a crucible weighed about 1 g of finely powdered air dried coal sample is placed inside an electric oven at 105–110°C for 1 h. After that, the crucible is taken out from the oven, cooled in a desiccator and weighed. Difference in the weight gives information about weight loss as moisture. weight loss % moisture = ×100 weight of coal (ii) Volatile matter: Dried matter of coal left in crucible (a) is covered with a lid and heated up to 950°C for 7 min in a muffle furnace. The crucible is cooled first in air, next in a desiccator and then weighed. Loss of weight is due to the volatile matter present. weight loss % volatile matter = ×100 weight of coal (iii) Ash: The residual sample in the crucible (b) is repeatedly heated and cooled (air and desiccator) up to getting content weight in muffle furnace around 700–750°C and the remaining residue is responsible for ash. weight of residue left % Ash = ×100 weight of coal (iv) Fixed carbon: Fixed carbon is determined by the following equation % Carbon = 100 − (% moisture + volatile matter + ash) Good quality of coal has more fixed carbon.

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Fuels and Combustion

3.9

3.7.2.2 Ultimate Analysis of Coal Elemental analysis of coal is done by ultimate analysis, and with this analysis carbon, hydrogen, nitrogen, oxygen, sulphur and ash content are determined based on the following procedure. (i) Carbon and hydrogen: In a combustion apparatus, accurately weighed 1–2 g of a coal sample is burnt in a oxygen current. The coal sample containing carbon and hydrogen is converted into carbon dioxide and water, and the formed gaseous products are absorbed by known weight of potassium hydroxide and calcium chloride tubes, respectively. From the weight difference of the tubes, percentage of carbon and hydrogen is determined as follows. C + O2 → CO2 (12 parts by mass of carbon gives 44 parts 12 44 by mass of carbon dioxide) 2KOH + CO2 → K 2 CO3 + H 2 O 1 H 2 + O2 → H 2 O (2 parts by mass of hydrogen gives 18 parts by mass of water) 2 2 18 7H 2 O + CaCl 2  → CaCl 2 ⋅ 7H 2 O % carbon =

weight increase in KOH tube × 12 × 100 weight of coal × 44

% hydrogen =

weight increase in CaCl 2 tube × 2 × 100 weight of coal × 18

(ii) Determination of Nitrogen content by Kjeldahl method: In a Kjedahl flask (long-necked flask), about 1g of accurately weighed powdered coal is heated with concentrated sulphuric acid and potassium sulphate as a catalyst. After getting a clear solution treated with excess KOH, liberated ammonia is distilled over and absorbed by known volume of standard acid solution. Unused acid is determined with standard NaOH by back titration. Nitrogen content in coal is calculated from the volume of acid used by liberated ammonia. % nitrogen =

volume of acid used × normality × 14 × 100 weight of coal

(iii) Sulphur: Sulphur content in coal is determined from the washings obtained in the determination of calorific value by the bomb calorimeter. During the determination of calorific value, entire sulphur present in coal is converted into sulphate. The washings are treated with barium chloride solution, and the sulphate is precipitated as barium sulphate, then it is filtered, washed and heated for obtaining a constant weight. % sulphur =

weight of BaSO 4 obtained × 32 × 100 233 × weight of coal taken in bomb calorimeter

(iv) Ash: Ash content is determined as in proximate analysis. (v) Oxygen: Oxygen content is determined by using the following equation % Oxygen = 100 − %(C + H + S + N + ash)

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3.10 Engineering Chemistry Solved Numerical Problems Based on Combustion of Fuel (i) A sample of coal was analysed as follows: Exactly 2.500 g was weighed into a silica crucible. After heating for one hour at 110 °C, the residue weighed 2.415 g. The crucible next was covered with a vented lid and strongly heated for exactly seven minutes at 950 ± 20 °C. The residue weighed 1.528 g. The crucible was then heated without the cover until a constant weight was obtained. The last residue was found to weight 0.245 g. calculate the percentage results of the above analysis. Solution Mass of moisture in coal sample = 2.500 – 2.415 = 0.085 g Mass of volatile matter = 2.451 – 1.528 = 0.887 g Mass of ash = 0.245 gm Percentage of moisture =

0.085 × 100 = 3.400% 2.5

Percent of volatile matter = Percent of ash =

0.887 × 100 = 35.48% 2.500

0.245 × 100 = 9.8% 2.5

Percent of fixed carbon = (100 – (3.4 + 35.48 – 9.80)) = 51.32% (ii) Calculate the mass of air needed for complete combustion of 5 kg of coal contain; C – 80%, H = 15%, O = rest. Solution 5 kg of coal contains: C = 4 kg; H = 0.75 kg; O = (5 – 4 – 0.75) kg = 0.25 kg \ Amount of air required for complete combustion of 5 kg coal = [5 × (32/12) + 0.75 × (16/2) – 0.25] kg × (100/23) = [13.333 + 6.000 – 0.25] kg × (100/23) = 82.97 kg (iii) A sample of coal was found to contain; C – 80%, H – 5%, O – 1%, N – 2% remaining being ash. Calculate the amount of minimum air required for complete combustion of 1 kg of coal sample. Combustion reaction

Wt. of air required

C + O2 → CO2 12 32 2H + 0.5O2 → H2O 2 16

800 g (32/12) = 2146 gm 50 g (16/2) = 400 gm Total = 2546 gm Less O in fuel = 10 gm Net O2 reqd. = 2536 gm

\ wt. of air reqd. = 2536 g (100/23) = 11026 gm = 11.026 kg.

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Fuels and Combustion

3.11

(iv) Calculate the weight and volume of air required for combustion of one kg of Carbon? Solution Carbon undergoes combustion according to the equation. C + O2 → CO2 12 32 Thus wt. of O2 required for combustion of 12 gm of C = 32 gm. 32 Hence weight of oxygen required by 1 kg of carbon = × 1 = 2.667 kg 12 100 \ wt. of air (containing 23% oxygen) required = × 2.667 = 11.59 kg 23 Now since 32 gm of oxygen occupies 22.4 litres at NTP \ 2.667 × 1000 gm of O2 will occupy =

22.4 × 2.667 × 1000 = 1866.9 L 32

So, volume of air (containing 21% oxygen) required =

100 × 1866.9 = 8890 Litres = 8.89 m3 21

(v) A gas has the following composition, by volume: H2 = 30%; CH4 = 5%; CO = 20%; CO2 = 6%; O2 = 5% and N = 34%. If 50% excess air is used find the weight of air actually supplied per m3 of this gas. [molecular weight of air = 28.97] Solution In one m3 of the gas Composition of components

Volume of O2 needed

H2 (30%) = 0.3 m CH4 (5%) = 0.05 m3 CO (20%) = 0.2 m3 O2 (5%) = 0.05 m3

0.3 × 0.5 = 0.5 m3 0.05 × 2 = 0.1 m3 0.2 × 0.5 = 0.1 m3 Total = 0.35 m3 Less O2 in fuel gas = −0.05 m3 Net O2 needed = 0.3 m3 = 300 L

100 150 Volume of air required for 1 m3 of gas using 50% excess air = 300 × × = 2142.8 L 21 100 Hence, weight of air actually supplied per m3 of the gas,  1 mol   28.97 gm  = 2142.8 L ×  ×  22.4 L   mol  = 2771 gm (vi) A gaseous fuel has the following composition by volume; H2 = 20%; CH4 = 5%; CO = 20%; CO2 = 5%; N2 = 45%. If 50% excess of air is used find the weight of air actually supplied per m3 of this gas?

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3.12 Engineering Chemistry Solution Volume of components in 1 m3 of gaseous fuel and O2 needed for combustion can be calculated as: Composition

Combustion equation

Volume of O2 needed

H2 = 0.20 m CH4 = 0.20 m3 CO = 0.20 m3

H2 + 0.5O2 → H2O CH4 + 2O2 → CO2 + 2H2O CO + 0.5O2 → CO2

0.20 × 0.5 = 0.1 m3 0.05 × 2 = 0.1 m3 0.2 × 0.5 = 0.1 m3 Total = 0.3 m3 = 300 L

∴ Volume of air required for 1 m3 of gas using 50% excess air 100 150 = 300 × × = 2142.8 L 21 100 Hence, weight of air actually supplied per m3 of gas = Volume ×

28.94 28.94 = 2142.8 × = 2768.4 L 22.4 22.4

(vii) Calculate volume of air required for complete combustion of litres of CO, given percentage of oxygen in air 21. Solution Combustion reaction volume of O2 needed CO + 0.5O2 → CO2 5 L × 0.5 = 2.5 L Hence, volume of air required 100 = 2.5 × = 11.9 L 21 (viii) A producer gas has following composition by volume: CH4 = 5%; CO = 30%; H2 = 20%; CO2 = 5%; N2 = 40%. Calculate the theoretical quantity of air required per cubic meter of the gas. Solution volume of component in 1m3 of gaseous fuel and O2 needed for combustion can be calculated as: Composition CH4 = 0.05 m CO = 0.3 m3 H2 = 0.2 m3

Combustion equation

Volume of O2 needed

CH4 + 2O2 → CO2 + 2H2O CO + 0.5O2 → CO2 H2 + 0.5O2 → H2O

0.05 × 2 = 0.1 m3 0.3 × 0.5 = 0.15 m3 0.2 × 0.5 = 0.1 m3 Total = 0.35 m3

\ Volume of air required for 1 m3 of gas = 0.35 ×

100 = 1.67 m3 . 21

(ix) A coal sample gave the following analysis: C = 66.2%; H = 4.2%; O = 6.1%; N = 1.4%; S = 2.9%; moisture = 9.7% and ash = 9.5%. If one kg of coal is burnt with 25% excess air, d etermine the quantity of products of combustion? Solution One kg of coal sample contains: C = 662 gm; H2 = 42 gm; S = 29 gm; O = 61 gm; H2O = 97 gm

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Fuels and Combustion

Combustion reaction

Weight of oxygen needed for combustion

C + O2 → O2

662 ×

H2 + 1/2O2 → H2O

42 ×

S + O2 → SO2

29 ×

32 = 1765.3 gm 12

3.13

Weight of products of combustion CO2 =

44 × 662 = 2427.3 gm 12

16 = 336 gm 2

H 2O =

18 × 4 L = 378 gm 2

32 = 29 gm 32

SO2 =

64 × 29 = 58 gm 32

Total = 2130.3 gm Less O2 in fuel = -61 gm Net O2 needed = 2069.3 gm

Hence, minimum weight of air required for complete combustion of 1 kg of coal = 2069.3 ×

100 = 8996.96 gm (1) 23

(Because the air has 23% (by oxygen weight)) And weight of air supplied for combustion using 25% excess air 125 = 11246.2 gm (2) 100 Since, total weight of products of combustion = 8996.96 ×

= Weight of [excess O2 + N2 + H2O + SO2 + CO2] (3)

\ We should first calculate individual weights of products. Now, weight of excess O2 = 25% of Net O2 used {equation (1)} 25 = × 2096.3 = 5173 gm 100 Weight of N2 = 77% of weight of air + weight of N2 present in fuel 77 = × 11246.2 + 14 = 8659.6 gm 100 Weight of H2O = 378 + 97 = 475 gm Weight of SO2 = 58 gm Weight of CO2 = 2427.3 gm \ Total weight of products of combustion = Weight of (excess O2 + N2 + H2O + SO2 + CO2) = 517.3 + 8659.6 + 475 + 58 + 2427.3 gm = 12137.2 gm = 12.137 gm (x) The percentage composition of a sample of bituminous coal was found to be as under: C = 75.4%; H = 5.3%; O = 12.6%; N = 3.2%; S = 1.3% and Ash = rest. Calculate the minimum weight of air necessary for complete combustion of 1 kg of coal and percentage composition of dry products of combustion by weight:

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3.14 Engineering Chemistry Solution Total weight O2 needed 32 16 32   =  C × + H × + S ×  gm  12 2 32  32 = 754 × + 53 × 8 + 13 × 1 = 2447.6 gm 12 Less O2 in coal = 126 gm \ Net O2 needed = 2321.7 gm So, minimum weight of air necessary for complete combustion = 2321.7 ×

100 = 10094.2 gm = 10.09 kg 23

Dry products of combustion 44 44 ×C = × 754 = 2764.7 gm 12 12 64 64 SO2 = ×S = × 13 = 26 gm 32 32

CO2 =

N2 = 77% of weight of air + in fuel =

77 × 10094.2 + 32 100

= 7804.5 gm Total weight of dry products of combustion = weight of (CO2 + SO2 + N2) = 2764.7 + 26 + 7804.5 = 10595.2 gm 2764.7 \ Percentage of CO2 = × 100 = 26.09% . 10595.2 Percentage of SO2 = Percentage of N 2 =

26 × 100 = 0.245% 10505.2

7804.5 × 100 = 73.66% 10595.2

(xi) The coal has following analysis: C = 54%; H = 6.5%; O = 3%; N = 1.8%; moisture = 17.3 and remaining is ash. This coal on combustion with excess of air, gave 21.5 kg of dry flue gases per kg of coal burnt. Calculate percentage of excess air used for combustion. Solution 1 kg of coal contains C = 0.54 kg: H = 0.065 kg; O = 0.03 kg; N = 0.018 kg Minimum weight of air required for combustion 32 16 100   =  0.54 × × 0.065 × − 0.03 × = 8.39 kg   12 2 23

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Fuels and Combustion

3.15

Weight of dry products of combustion 44 = 19.8 kg 12 77 N 2 = 0.018 × 8.39 × = 6.478 kg 100

CO2 = 0.54 ×

\ Total weight of dry products combustion = 1.98 + 6.478 = 8.458 kg Given, the actual weight of dry flue gases is 21.5 kg. so balance must have come from excess air = 21.5 - 8.458 = 13.42 kg 13.042 Hence percentage of excess air = × 100 = 155.45% 8.39 (xii) The percentage composition of a sample of coal by weight was found to be C = 76%; H = 5.2%; O = 12.8%; N = 2.7%; S = 1.2%. the remaining being ash. Calculate the minimum: (a) weight, and (b) volume at NTP of air necessary for complete combustion of 1 kg of coal. Also calculate percentage composition of dry products by weight, if 50% excess air is supplied. Solution 1 kg of coal contains C = 760 gm: H = 52 gm; S = 12 gm; O = 128 gm; N = 27 gm \ Net O2 needed for combustion = (O2 needed for combustion)-(O2 is fuel) 16 32  32  =  × 760 + × 52 + × 12 − (128) = 2326.7 gm  12  2 32 Now, weight of air necessary for complete combustion of 1 kg of coal = 2326.7 ×

100 = 10115.9 gm = 10.116 kg 23

And volume of air necessary for complete combustion of 1 kg of coal = 10115.9 gm ×

22.4 L 1 m3 × = 7.83 m3 28.94 gm 1000 L

Weight and percentage of dry products of combustion are calculated below: CO2 =

44 × 760 = 2786.7 gm 12

SO2 =

64 × 12 = 24 gm 32

N 2 = 27 gm (in fuel) +

77 150 × 10116 × = 11710.9 gm 100 100

O2 = Minimum weight of O2 ×

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50 50 = 2326.7 × = 1163.7 gm 100 100

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3.16 Engineering Chemistry Total weight of dry products of combustion = weight of (CO2 + SO2 + N2 + O2) = 2786.7 + 24 + 11710.9 + 1163.4 = 15684.97 gm \ Percentage of CO2 =

2786.7 × 100 = 17.77% 15684.97

Percentage of SO2 =

24 × 100 = 0.153% 15684.97

Percentage of N 2 =

11710.9 × 100 = 74.7% 15684.97

Percentage of C2 =

1163.4 × 100 = 7.417% 15684.97

3.7.3 Metallurgical Coke Coke used for metallurgy is called metallurgical coke, and it should have the following good characteristics. (i) Purity: Low moisture content keeping down the heating expenses. Ash content hinders the heating, forms slag and also consumes excess coke for removal of ash. The sulphur and phosphorous produce undesirable products like SO2, P2O3, P2O5 etc. Which affect the quality of coke, and also sulphur make the coke brittle. Hence coke should have as low as possible moisture, ash, phosphorous and sulphur contents. (ii) Calorific value: Coke should have a high calorific value. (iii) Strength: It should be quite compact, strong, hard to withstand abrasion as well as pressure in furnace. (iv) Porosity: For complete combustion at high rate, coke should be porus, due to presence of pores oxygen can easily contact with carbon. (v) Size: If coke is too big in size, uniformity of heating is not maintained, and if it is too small choking is observed. Hence, the size of the metallurgical coke should be medium. (vi) Cost availability and transportation: Coke should be easily available with cheap rate nearer the metallurgical plant, therefore, with this the transportation cost is also reduced. (vii) Combustibility: The combustibility of the coke mainly depends on nature of coal, carbonization temperature, reaction temperature, etc. Further, cokes obtained by high temperature carbonization process are less combustible when compared to coke obtained by low temperature carbonization at a given temperature. All cokes have equal reactivity at 800–900°C temperature. The rate of combustion depends on the rate of oxygen supply about 1000°C. Coke should burn easily. (viii) Reactivity of steam: Coke obtained from non-caking coals is more reactive to steam when compared with caking coals. Reactivity to steam of coke is directly proportional to reaction temperature and inversely proportional to carbonization temperature. Especially, the coke used for manufacture of water gas must be reactive to steam.

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Fuels and Combustion

3.17

3.7.4 Manufacture of Metallurgical Coke The coke, for metallurgical purposes, is mainly manufactured by two methods. They are (1) Beehive oven and (2) Otto Hoffman’s by-product oven method. (i) Beehive oven: Schematic representation of Beehive oven is shown in Figure 3.3. This is the cheap and earliest method for manufacturing of metallurgical coke. The Beehive oven is the domeshaped structure of bricks, with 4 m width and 2.5 m height. It is having two openings, and these can be opened and closed as desired. Thus, coal is charged from the top opening, and air supply as well as coke discharge from side opening is used. Through the top opening, coal is charged about 0.6 m deep layer, air is supplied from the side opening and the coal ignited. For slow carbonization, combustion is allowed to proceed gradual diminish supply of air, and it will take to complete 3 to 4 days from the top to bottom layer and the volatile matter escapes inside the partially closed door. After completing and carbonization, the hot coke quenched with water and raked out through the side door, leaving the oven hot to start the next charge batch carbonization. The yield is 80 per cent of the charged coal. Many such ovens are arranged in series, and with this waste heat is utilized for heating. Hence, it saves energy, reduces the pollution and is economically beneficial. Coal charging door Refractor lining Heat radiated from the roof to the coal bed

Zone of combustion Coal

2.5 m

0.6 m

Door for air supply or coke discharging

Figure 3.3 Beehive coke oven (ii) Otto Hoffman’s by-product oven: Schematic representation of the modern by-product coke oven, which is developed by Otto Hoffman is shown in Figure 3.4. It is mainly useful in (i) increased thermal efficiency of carbonization and (ii) recovery of valuable by-products like ammonia, coal gas, benzol air, tar etc., It is heated externally by coal gas produced itself or blast furnace gas or producer gas, and mostly heating is done by heat economy of regenerative system, i.e., utilization of flue gases for heating. The oven consists of a number of narrow silica chambers about 10 to 12 m long, 3 to 4 m high and 0.40 to 0.45 m width. These chambers are erected side by side vertically; further, flues in between them form a sort of battery. Each chamber is provided at the top with a charging hole, at the end of chamber a gas off-take and refractory lined cast iron door for discharging coke. A finely crushed coal is introduced through the charging holes, closed tightly on both the ends to prevent air access. The oven is heated to 1200°C by employing a regenerative principle, with burning of producer gas. During combustion, produced flue gases pass towards sensitive checker brick work until the temperature raises about 1000°C before escaping to chimney. The flow of heating gases is reversed, to serve in the preheat of inlet gases and the cycle goes on. The heating process is continued up to 11 to 18 h, till the carbonization and evolution of volatile matter ceases completely. After complete carbonization, a massive ram pushes the red hot coke into a truck and subsequently quenched.

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3.18 Engineering Chemistry Chambers loaded with coal l

a Co

Coal gas

Coal bunker Coal gas

Coal charging cars

Tar Coke chamber

Ram (Coke pusher)

Gas burns here

Outgoing waste gases heat regenerators before escaping to chimney

Regenerators

Coke oven with regenerators Waste gases to chimney

Space between chambers for flow of burnt gases Hot regenerators

Air Producer Waste gases to chimney gas

Figure 3.4 Otto Hoffman’s by-product coke oven with regenerators

3.8 Liquid Fuels Liquid fuels are those which are combustible, energy-generating substances and play vital role in transportation and economy. Most widely used liquid fuels are derived from fossil fuel/petroleum/ crude oil. Some important liquid fuels are petrol, kerosene, diesel, etc.

3.8.1 Petroleum Refining Petroleum is a complex mixture of organic liquids (hydrocarbons) also known as crude oil or fossill fuel. It is formed from the fossilized dead plants and animals by exposure to heat and pressure in the Earth’s crust, and was formed millions of years ago. It is a viscous dark coloured, foul-smelling liquid along with water and soil particles. Hence, it is necessary to separate these hydrocarbons into useful products, and this process is known as fractional distillation. In this process, products are separated depending on boiling points, known as refining of petroleum, and the plant set-up used here are oil refineries as shown in Figure 3.4. Refining of petroleum involves the following 3 steps. Step I: Separation of water by Cottrell’s method: Petrol or crude oil is the emulsion of oil and salt water, and these colloidal water droplets coalesce to form large drops which can separate out from oil when the crude oil is sent through two highly charged electrodes. Step II: Removal of sulphur compounds: crude oil is treated with copper oxide, sulphur reacts with copper to form copper sulphide precipitate, which is removed by filtration Doctors sweetening process: The process was described by G. Wendt and S. Diggs. Here, crude oil is treated with sodium plumbate, i.e., doctors solution, converts mercaptans in sour gasoline into disulphide. RSH

Mercaptan from Gardine

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S in presence of NaOH Na 2 PbO2 powdered  → R − S − S − R + PbS + 2NaOH

(sodium plumbate)

(alkyl disulphide)

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Fuels and Combustion

3.19

Gases

Tray Chimney Down spot Loose cap

Petroleum ether Gasoline Naphtha Kerosene Diesel oil Lubrication oil

Steam Crude oil

Fumace at 400°C

Heavy oil

Figure 3.5 Fractional distillation of crude petroleum Step III: F ractional distillation: In an iron retort, the crude oil is heated to about 400–430°C. Here, all volatile matter are evaporated, components which are not volatile like tar and asphalt are settled at the bottom of the column. The hot vapours are then passed through a distillation column, shown in Figure 3.5. The distillation chamber is a steel cylindrical tube about 31 m height and 3 m in diameter, and inside, the chamber trays are fitted at short distances. Every tray is having many holes and an up going short tube with a bubble cap. At different heights of chamber, the vapours go up, begin to cool and condense in fractions. Fractions which are having higher boiling point condenses first and lower boiling fractions one after other. Various products obtained in distillation are given in Table 3.1. Table 3.1 Fractions by distillation of crude Approx. composition in terms of hydrocarbon containing C atoms

Uses

Name of fraction

Boiling range

1. Uncondensed gas.

Below 30 °C

C1 to C4 (such as ethane, propane, isobutane)

As domestic or industrial fuel under the name L.P.G. (liquefied petroleum gas).

2. Petroleum ether.

30–70 °C

C5–C7

As a solvent. (Continued )

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3.20 Engineering Chemistry Table 3.1 (Continued) Approx. composition in terms of hydrocarbon containing C atoms

Name of fraction

Boiling range

Uses

3. Gasoline or petrol or motor spirit.

40–120 °C

C5–C9 (calorific value = 11,250 kcal/kg)

As motor fuel, solvent and in dry cleaning.

4. Naphtha or solvent spirit.

120–180 °C

C9–C10

As solvent and in dry cleaning.

5. Kerosene oil.

180–250 °C

C10–C16 (calorific value = 11,000 kcal/kg)

As an illuminant, jet engine fuel and for preparing laboratory gas.

6. Diesel oil or fuel oil or gas oil.

250–320 °C

C10–C18 (calorific value = 11,000 kcal/kg)

Diesel engine fuel.

7. Heavy oil. This on refractionation gives: (a) Lubricating oil (b) Pertoleum jelly. (Vaseline) (c) Grease (d) Paraffin wax.

320–400 °C

C17–C30

For getting gaso-line by cracking process. As lubricant As lubricant and in cosmetics and medicines. As lubricant In candles, boot polishes, wax paper, tarpolin cloth, etc.

8. Residue may be either: (a) asphalt or (b) Petroleum coke.

Above 400 °C C30 and above

Water-proofing of roofs and road making. As a fuel and in moulding are light rods.

3.8.2 Important Petroleum Products and their Uses (i) Gasoline (or) petrol (or) motor spirit: In North America, gasoline is often shortened as gas, while petrol is the common name in the United Kingdom. It is a transparent petroleum derived oil obtained between 40 to 120°C as mixture of hydrocarbons C5H12–C8H8. Its calorific value is about 11,250 kcal/kg, with 84 per cent of carbon, 15 per cent of hydrogen and 1 per cent of nitrogen, sulphur and oxygen as its composition. It is highly volatile inflammable oil, primarily used as a fuel for internal combustion engines of automobiles. (ii) Kerosene oil: Kerosene is the fraction obtained between 180 to 250°C, as a mixture of hydrocarbons C10H22–C16H34 in petroleum distillation. Its calorific value is about 11,000 kcal/kg, with 84per cent of carbon, 16 per cent of hydrogen and less then 0.1 per cent of sulphur as its composition. It is widely used fuel for cooking, also used as an jet engine fuel and for making oil gas. (iii) Diesel oil: In a petroleum distillation, it is a fraction obtained between 250 to 320°C as a mixture of hydrocarbons of C15H32–C18H38. Its calorific value is about 11,000 kcal/kg, and is mainly used as a diesel engine fuel.

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3.21

(iv) Liquefied petroleum gases (LPG): This is obtained from cracking of heavy oils (or) natural gas and named as bottled gas. It consist of n-butane, isobutene, butylene, propane and less ethane, and supplied under pressure containers with the trade name of Indane gas, Bharat gas, etc. LPG is dehydrated, desulphurized and added trace amounts of mercaptans for giving warning of gas leak, and its calorific value is about 27,800 kcal/m3. It is widely used as a domestic, industrial and motor fuel.

3.9 Synthetic petrol Petrol can be synthesized by the following methods. (i) Cracking (ii) Fischer–Tropsch method and (iii) Bergius method

3.9.1 Cracking The process of breakdown of high molecular weight hydrocarbons of high boiling points into simple, lower molecular weight hydrocarbons of low boiling points is known as cracking. cracking

C10 H 22 → C5 H12 + C5 H10

Example:

n-pentane

Decane

pentene

With these we can prepare different fuels having high quality. Cracking is mainly two types: thermal and catalytic cracking. 3.9.1.1 Thermal Cracking In this cracking heavy oils are subjected to high temperature and pressure in the absence of catalyst. In this cracking, the bigger hydrocarbon molecules break down to give smaller paraffins and olefins. Mechanism of Cracking Process Cracking processes invoke free radical and carbonium ion intermediates. Thermal cracking mainly goes through the free radical mechanism. In this mainly of three steps they are as follows: Example: Thermal cracking of nonane. Initiation: Nonane undergoes homolytic cleavage at high temperature to give free radicals. •

High temp

CH 3 − (CH 2 )7 − CH 3  → 2CH 3 − (CH 2 )3 − C H 2 Propagation: The formed free radicals undergo further fissions up to thermally more stable radical is formed. •

•

CH 3 − CH 2 − CH 2 − CH 2 − C H 2 → CH 3 − CH 2 − C H 2 + CH 2 = CH 2 •

•

CH 3 − CH 2 − C H 2 → C H 3 + CH 2 = CH 2 •

CH 3 C H 2 → H 2 C = CH 2 + H i

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3.22 Engineering Chemistry Termination: Coupling of unstable free radical intermediates gives final products in termination reaction. •

•

C H 3 + C H 3 → CH 3 − CH 3 •

•

H 3 C − C H 2 + C H 2 − CH 3 → H 3 C − CH 2 − CH 2 − CH 3 •

H 3 Ci + C H 3 → H 3 C − CH 3 •

H 3 C − C H 2 + H i → H 3 C − CH 3 3.9.1.2 Catalytic Cracking In catalytic cracking, higher molecular weight hydrocarbons breakdown in the presence of catalyst like alumina (or) aluminium sulphate via carbonium ion intermediate. Here, quality and quantity of gasoline can be increased, and it is mainly of two types. They are as follows: (i) Fixed bed catalytic cracking (ii) Moving bed catalytic cracking Mechanism of Catalytic Cracking This reaction proceeds via carbonium–ion intermediates. (i) Fixed-bed catalytic cracking: A simple sketch of fixed bed catalytic cracking is shown in Figure 3.6. Here, heavy oil charge is passed through a pre-heater, having a temperature of about 425– 450°C. The formed hot vapours of oil is passed over a fixed bed of catalyst chamber also having the temperature of about 425–450°C. Catalyst

Cracked vapour

Cooler

Stabilizer Gases

Vapour Gasoline Heavy oil charge

+ Heater − Catalyst Pre-heater chamber (425−450°C) (425−450°C)

Fractionating coloumn

Heavy oil Gasoline + Some dissolved gas

Figure 3.6 Fixed-bed catalytic cracking In a catalytic chamber, 40 per cent of oil is converted into petrol and 2–4 per cent of carbon formed is absorbed on the catalyst bed. Catalyst stops function after 8–10 h, and due to carbon deposition it deactivates. This is re-activated by burning off the deposited carbon, During re-activation, the vapours are directed through another catalyst chamber. Cracked vapours enter into the fractionating column from the catalyst chamber, and different gases are cooled and collected. (ii) Moving-bed catalytic cracking: A schematic representation of a moving bed catalytic cracking is shown in Figure 3.7. Here, feed oil is passed through a pre-heater, oil vapours formed here

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along with very finely powdered catalyst are passed to a reactor, which maintain about 500°C temperature. The cracked oil vapours are then passed through fractionating column, and heavy oil is separated. Formed vapours are sent to a cooler, gasoline condenses along with gases, and is separated from gases as a purified petrol. Flue gases Cyclone

Catalyst regenerator

600°C

Regenerator Catalyst

Cracked vapour Light fraction

Cooler Gases Gases

500°C

Stabilizer Reactor

Feed oil

Spent catalyst

Catalyst + oil

Heavy oil Fraction Gasoline matching + gas coloumn Air

Gasoline

Blower

Figure 3.7 Moving-bed type catalytic cracking

3.9.2 Fischer–Trapsch Method Oven-heated coke is mixed with hydrogen and passed steam through it, and water gas (CO + H2) is formed. It is purified by passing through first Fe2O3, here H2S is removed, next a mixture of Fe2O3 + Na2CO3, removes organic sulphur compounds. The purified gas is then compressed to 5–25 atm and is sent through a converter containing catalyst. Catalyst is the mixture of 100 parts of cobalt, 8 parts of magnesia, 5 parts of thoria and 200 parts of keiselgur at 200–300°C temperature. A mixture of saturated and unsaturated hydrocarbons is formed. nCO + 2n H 2  → Cn H 2n + nH 2 O unsaturated hydrocarbon

nCO + (2n + 1) H 2  → Cn H 2n + 2 + nH 2 O saturated hydrocarbon

This reaction is highly exothermic. Hence, formed hot gaseous mixture is sent to a cooler. Here, liquid like crude oil is formed, and passed through a fractionating column. From the column, petrol and heavy oil are formed. Cracking of heavy oil gives and petrol. Schematic diagram of Fisher-Tropsch method is shown in Figure 3.8.

3.9.3 Bergius Method A paste of finely powdered low ash coal, heavy oil and tin or nickel oleate (catalyst) is heated with hydrogen at 450°C temperature, and 200–250 atm pressure for about 1.5 h. Here, hydrogen reacts with coal to give saturated hydrocarbons, these are send to condense. Liquid like crude is formed and sent to fractionating column. From the column petrol, middle oil and heavy oil are formed. Heavy oil is used further for making paste with fresh coal. The middle oil is hydrogenated in presence of a solid catalyst in vapour phase to give petrol. The schematic diagram of Bergius process is shown in Figure 3.9.

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3.24 Engineering Chemistry Catalyst (Co + Th + MgO) + Keiselguhr) Cooler Fe2O3

Fe2O3

+ Gasoline

Na2CO3 Water gas (CO + H2)

Cracking Heavy oil

Purification of gas

Compressor (5-25 atm)

Gasoline

Fractionating column

Figure 3.8 Fischer–Trapsch method Powedered coal Heavy oil

Catayst (Sn or Ni oleate)

Gases

Condenser

Gasoline Paste

Middle oil H2

Gasoline

Heavy oil

Convertor at 450°C and 200-250 atm

Crude oil

Fractionating coloumn

Figure 3.9 Bergius method

3.10 Power alcohol Power alcohol is one of the most important non-petroleum fuels. The first four aliphatic alcohols, methanol, ethanol, propanol and butanol, can be synthesized chemically or biologically and used as a fuel for internal combustion engines. These are not used as a prime fuel, but used in blends as additives. Chemical formula of power alcohol is CnH2n+1 OH.

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3.10.1 Manufacture of Power Alcohol Methanol can be prepared from biomass. Ethanol is commonly prepared from various biological organic substances through fermentation process. However, widely it is manufactured from molasses. It is a viscous semisolid material, left after crystallization of sugar from sugar cane juice. It is a mixture of sucrose, glucose and fructose. The molasses are diluted with water, to reduce sugar concentration from about 50–60 per cent to 10–12 per cent. Nutrients like ammonium sulphate, ammonium phosphate, and some amount of sulphuric acid is added to maintain pH value around 4-5. Right proportions of yeast are added and maintain the temperature of about 30°C. The invertase enzyme of yeast converts entire sucrose into glucose and fructose. C12 H 22 O11 + H 2 O Invertase  → C6 H12 O6 + C6 H12 O6 30° C

Sucrose

Glucose

Fructose

The zymase enzyme of yeast converts entire glucose and fructose into ethyl alcohol and releases carbon dioxide. During this process CO2 produces lot of froth, hence this process is known as fermentation process. C6 H12 O6 Zymase   → 2C2 H 5 OH + 2CO2 (g) 30° C Glucose/fructose

Ethyl alcohol

The fermentation process may be completed in about 36–38 h. Depending on the concentration of alcohol, it is named as wash or rectified spirit or absolute alcohol. Wash: The fermented liquid containing 18–20 per cent of alcohol is known as wash. Rectified spirit: Fractional distillated wash contains 90–95 per cent alcohol, and it is known as rectified spirit. Absolute alcohol: The rectified spirit is digested with lime for about 2 days and then distilled to get 100 per cent alcohol which is known as absolute alcohol. Advantages (i) These are prepared from waste, hence it is a good non petroleum alternative source of energy and also reduces the pollution. (ii) It can burn completely, thereby increasing combustion efficiency. (iii) It has an octane value of about 90, but petrol is having about 60–70. When alcohol mixes with petrol, it tends to increase the octane rating, and these blended petrol possesses better antiknock property and reduces the carbon monoxide emission. (iv) Petrol is blended with alcohol, and it can absorb traces of moisture. Disadvantages (i) Alcohol may cause corrosion due to easy oxidation with acids. (ii) Due to low calorific value of alcohol, more fuel is required for each mile driven. (iii) Particularly at low temperatures, alcohol is difficult to atomize, due considerable surface tension.

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3.26 Engineering Chemistry

3.11 Knocking Knocking is the metallic sound produced by a spark ignition petrol engine under certain conditions. The following terms can explain the knocking in better way. (i) Ignition temperature: It is the minimum temperature at which the combustion is self-supporting. This is also referred to as spontaneous ignition temperature (SIT). (ii) Compression ratio (CR): The power output and efficiency of an IC engine depends on a factor called CR. It is defined as the ratio of gaseous volume (V1) in the cylinder at the end of suctionstroke to the volume (V2) at the end of compression-stroke of the piston. V As V1 > V2 ⇒ CR = 1 > 1 V2

The CR obviously indicates the extent of compression of fuel–air mixture by the piston. The fuel–air mixture gets heated to a temperature greater than its ignition temperature as a result of compression. This leads to spontaneous combustion even before sparking. It is also possible that the last portion of the fuel–air mixture undergoes self-ignition after sparking. It is due to the heating and compression of the unburned fuel, by the spreading flamefont sweeping across the cylinder. The resulting shock wave dissipates its energy by hitting the cylinder walls and the piston. In view of the characteristic rattling sound emitted, this is called knocking. The CR at which fuel tends to knock is called critical CR. To summarise: With the increase in CR, the efficiency of IC engine also increases but after critical CR, tendency to knock also increases. Consequences of knocking: (a) Decreased power output (b) Mechanical damage by overheating of the cylinder parts.

Probable mechanisms of chemical reactions that lead to knocking are the following: (c) Free radical chain reaction leading to cracking and oxidation of the hydrocarbons is probably the mechanism of chemical reactions that lead of knocking. Factors on which knocking depend are the following (1) Engine design (2) Running conditions and (3) Chemical structure of the fuel hydrocarbons. For instance: (d) Knocking tendency decreases in the following order: n-alkanes > mono-substituted alkanes > cycloalkanes > alkenes > poly-substituted alkenes > aromatics. And for straight chain hydrocarbons, the tendency to knock increases with molecular weight and boiling point. Example: n-hexane > n-pentane > n-butane. Aromatic hydrocarbons have higher anti-knocking properties than paraffins and olefins.

3.12 Diesel Engine, Cetane and octane Number In the diesel engine, air is first drawn into the cylinder and compressed to a pressure of about 500 psi (3.52 × 105 kg/m2). This compression is accompanied by a rise in temperature to about 500 °C.

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Towards the end of the compression, stroke is injected in the form of finely divided spray into air in the cylinder heated to about 500 °C by compression. The oil absorbs the heat from the air and it ignites spontaneously as it attains ignition temperature. This raises the temperature and pressure. The piston is pushed by expanding gases in the power stroke. In a diesel engine, combustion of fuel is not instantaneous, as the ignition delay is caused. Ignition delay is the interval between the start of fuel injection and its ignition. This is due to the time taken for the vaporization fuel droplets and attaining of the vapour to its ignition temperature. It depends on the (a) engine design; (b) efficiency of mixing of the spray and air; (c) the injector design; and (d) mostly on the chemical nature of the fuel. Example: Ignition delay is shorter for paraffinic fuel than that of olefinic, naphthalenic and aromatic fuels. If the ignition delay is long, it will lead to fuel accumulation in the engine even before the ignition. When ignited, an explosion results as the combined effect of increased temperature and pressure. This is responsible for diesel knock. The diesel fuel should have a SIT less than the temperature produced by compression. As the temperature to which air can be heated by compression is limited by various constraints, it is desirable to have fuels with short ignition delay but the ignition delay must be long enough for the compression stroke to be completed. In order to grade the diesel fuels, cetane rating is employed. Cetane number: It is used for diesel engines to measure the ease of with which a fuel will ignite under compression Diesel fuel

Cetane no.

Cetane C16H34 (or CH3 (CH2)14–CH3, n-hexadecane) X-methyl naphthalene

100 0

Remarks Very short ignition delay Longer ignition delay

CH3

Cetane number of fuel primarily depends on the nature and composition of its hydrocarbons. For instance, consider the following series: n-alkanes > cycloalkanes > alkenes > branched alkanes > aromatics (i.e. cycloalkanes): (i) ignition delay increases from left to right (ii) ignition quality increases from right to left (iii) cetane no. increases from right to left As straight chain alkanes such as n-cetane have low ignition delay (high ignition quality) and ignite readily on compression, while aromatics do not ignite readily on compression, so that high cetane number fuels eliminate diesel knock. The cetane number of diesel fuel may be raised by the addition of pre-ignition dopes such as alkyl nitrites such as ethyl nitrite, amylnitrite, etc., 2,2,4,4,6,6,8, 8-hepta methyl nonane (HMN). CH3 CH3

C CH3

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CH3 CH2

C CH3

CH3 CH2

C CH3

CH3 CH2

CH3

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3.28 Engineering Chemistry With a cetane rating of 15 is now considered as the low-quality diesel in the view of its easy availability and purity. On the revised scale (HMN reference), the cetane number (CN) represents the % cetane, in the blend with HMN plus 15/100 of the % HMN. Thus, a blend of 50% cetane and 50% HMN has a following cetane rating: 15 50 + × 50 = 57.5 100 Octane number: The resistance offered by gasoline to knocking cannot be defined in absolute terms. Itis generally expressed on an arbitrary scale, known as octane rating. Fuel n-Heptane Iso-octane

Octane number 0 100

Characteristics Knocks severely High resistance to knocking

The % of iso-octane in the n-heptane iso-octane blend which has the same knocking characteristics as the gasoline sample under the same set of conditions is known as octane number. Additives for improving anti-knock properties: Tetra ethyl lead (TEL) and diethyl telluside (C2H5)2Te are the most commonly used additives. TEL gives rise to Pb of PbO during combustion. These particles act as free-radical chain inhibitors as they arrest the propagation of the explosive chain reactions responsible for knocking. The efficiency of TEL decreases in the presence of sulphur hence desulphurised gasoline is preferred. Pb and PbO2 decrease engine life hence they must be removed along with exhaust gases by adding ethylene dibromide. Pb, PbO2 + C2H2Br2 → PbBr2 Because PbBr2 formed is volatile its escape into atmosphere. But pollution problem still exists. Another cause of pollution is incomplete combustion leading to the formation of CO, NO, NO2, SO2, SO3, etc. Hence, catalytic converters based on Pt are employed which will catalyse combustion reaction to completion. Example: CO–CO2. But Pt is poisoned by Pb, so unleaded petrol should be used. Benzene is added for decreasing knocking. Since benzene is carcinogenic, very low concentration of benzene should be used.

3.13 Gaseous fuels Important gaseous fuels are natural gas, producer gas, water gas, coal gas, bio gas, etc.

3.13.1 Natural Gas Natural gas obtained along with petroleum in oil wells is called wet gas. It is purified and removed. Propane, propene, butane, butene, etc. are used for preparing LPG. If the gas is associated with crude oil, it is called dry gas. It is having some of the objectionable ingredients like water, H2S, N2, CO2, etc. and hydrocarbons like propane, butane, propene, butene, etc. are removed. Composition Natural gas consists of 70–90 per cent of methane, 5–10 per cent of ethane, 3 per cent of hydrogen and rest of carbon monoxide and carbon dioxide, approximately. Calorific value is about 12,000–14,000 kcal/m3.

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Uses (i) Used as a domestic fuel, also conveyed over large distances through pipelines. (ii) Used as a raw material in carbon block manufacture. (iii) Used for manufacture of different synthetic chemicals. (iv) Used in the preparation of synthetic products by microbiological fermentation of methane. (v) Used in the preparation of compressed natural gas. (vi) Due to less pollution, it is a good substitute for petrol and diesel.

3.13.2 Producer Gas (or) Suction Gas Composition Producer gas is the mixture of about 20–22 per cent carbon monoxide (CO), 11–13 per cent carbon dioxide (CO2), 20–22 per cent hydrogen (H2), 2.5–3.5 per cent methane (CH4) and 40–42 per cent nitrogen (N2). Hence main composition is CO + N2. Manufacture Air is passed through a red hot coal or coke in a gas producer, and maintained temperature is about 1100°C. Producer gas is formed with oxidation and reduction reactions. Initially, oxidation of carbon gives carbon monoxide and carbon dioxide. C + O 2  → CO2 + Energy   exothermic reaction C + 1/ 2 O2  → CO + Energy  Reduction reaction gives producer gas:  CO2 + C  → 2CO + Energy  C + H 2 O  → CO + H 2 + Energy  endothermic reaction C + 2H 2 O  → CO2 + 2H 2 − Energy  Formed gas is distilled and purified. The calorific value of producer gas is about 1300 kcal/m3. Uses (i) Used in the manufacture of steel, glass, etc. for heating of open-hearth furnace. (ii) Used in the manufacture of coke and coal gas for heating of muffle furnace. (iii) Used as a reducing agent in metallurgical operations.

3.13.3 Water Gas (or) Blue Gas Composition Water gas is the mixture of carbon monoxide (40–42 per cent), hydrogen (50–52 per cent), nitrogen (3–4 per cent and carbon dioxide (3–4 per cent).

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3.30 Engineering Chemistry Manufacture Steam and little air are passed alternatively through a red hot coal or coke in a reactor maintained at about 1000°C temperature and water gas is formed in the following reactions. C + O 2  → CO2 + Energy   exothermic reaction 2C + O2  → 2CO2 + Energy  C + H 2 O  → CO + H 2 − Energy (endothermic reaction) The calorific value of water gas is about 2800 kcal/m3. Uses (i) Used as an illuminating gas, fuel gas, source of hydrogen gas etc. Carbonated water gas: It is a mixture of producer gas and hydrocarbons. Calorific value is about 4500kcal/m3, and used for illuminating and heating purpose. Semi-water gas: It is a mixture of water gas and producer gas. Calorific value is about 1700 kcal/m3. Used as a fuel and a source of N2 and H2 in the manufacture of ammonia.

3.13.4 Coal Gas Coal gas is mainly used as an illuminant in cities and towns; hence, it is known as town gas or illumination gas. Composition It is a mixture of carbon monoxide (27–29 per cent), carbon dioxide (2.4 per cent), hydrogen (16–18 percent), nitrogen (49–51 per cent) and methane (0.5–1 per cent). Manufacture It is manufactured by destructive distillation of coal in the absence of air, at about 1300°C temperature. 1300° C Coal In  → Coal gas ↑ the absence of air

The calorific value of coal gas is about 4900 kcal/m3. Uses (i) It is used as a fuel and illuminant. (ii) Used for maintaining reducing atmosphere in metallurgical operations.

3.13.5 Biogas Composition Biogas is the mixture of methane (50–60 per cent), carbon dioxide (30–40 per cent), hydrogen (5–10 per cent), nitrogen (2–6 per cent) and trace amount of hydrogen sulphide.

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Manufacture It is produced by the degradation of biological matter like animal dung, poultry waste, vegetable waste, waste paper, plant waste, human excreta, birds’ excreta, etc. by the anaerobic bacterial action in the absence of free oxygen. Uses (i) Used for cooking food. (ii) Used as a fuel to run engines. (iii) Used as an illuminant.

3.14 Flue gas analysis by Orsats apparatus Flue gas is the mixture of CO2, CO and O2 gases, exhausted from the combustion chamber. Analysis of flue gas gives an idea about efficiency of combustion. Suppose the flue gas contains considerable amount of CO, it indicates incomplete combination and short supply of oxygen, and this will lead to wastage of fuel. If the flue gas contains considerable amount of oxygen, this indicates excess supply of oxygen and results in loss of heat. With the help of Orsat’s apparatus flue gas analysis is carried out, as is shown in Figure 3.10. The setup consists of a horizontal tube, with a three-way stopcock at one end and another end is connected with a graduated burette. For maintaining constant temperature during the experiment, the burette is surrounded by a water jacket. The burette is connected as a set of three absorption bulbs in a series, through a separate stopcock. The lower end of the burette is further connected to a water reservoir through a rubber tube. The water level in the burette can be changed by raising or lowering the reservoir water. One end of the tube, which is connected to a three-way stopcock, is further connected to a U-tube. For drying flue gas and avoiding the incoming smoke particles, the U-tube is packed with fused CaCl2 and glass wool. Among the three absorption bulbs, first bulb has potassium hydroxide solution and absorbs only CO2. The second bulb contains alkaline pyrogallic acid absorbs only O2 and CO2. The third bulb has ammonical cuprous chloride and can absorb CO2, O2 and CO. For proper analysis of flue gas, first it is passed Stop cock

Stop cock

Flue gas

Water reservoir 3

Fused CaCl2 + Glass wool

Amm cyprous chloride

Alk pyrogallic acid

Absorption bulbs

KOH

Graduated burrete

Water jocket

Rubber tubing

Figure 3.10 The orsats apparatus

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3.32 Engineering Chemistry through potassium hydroxide containing bulb, and here CO2 is absorbed. Then, it is passed through alkaline pyrogallic acid containing bulb, and here O2 is absorbed and it can also absorb CO2, but already it is removed by KOH. Finally, flue gas is passed through the third bulb containing ammoniacal cuprous chloride, and here CO is absorbed; however, it can absorb CO2 and O2 also but these are already removed. The entire apparatus is thoroughly cleaned, the steppers are greased, tested for air tightness, the absorption bulbs are filled with their respective solution and the stopcocks are closed. The water reservoir and water jacket are filled with water, and air is excluded from the burette by the raising of reservoir water level till the burette is completely filled with water. For the exclusion of air, the three-way stopcock is opened, next the lowering of water level is done and the fuel gas supply is connected to the three-way stopcock. Further, 100 ml of the flue gas is carefully sent to the burette with closing of the three-way stopcock. The fuel gas is forced through the first bulb by opening its stopcock and raising the water level in the reservoir. Here, potassium hydroxide absorbs the CO2 flue gas is sent repeatedly 2 or 3 times to the first bulb for complete absorption of CO2. The remaining gas is taken back in the burette and the stopcock of the first bulb is closed. The levels of water in the reservoir and burette are equalized and decreasing volume of gas is noted. This decrease in volume gives the volume of CO2 in 100 ml of flue gas. Similarly, the volumes of O2 and CO are determined by passing the flue gas through the second and third bulbs, respectively. The remaining gas in the burette after absorption of CO2, O2 and CO is nitrogen. The decrease in volume of flue gas by first bulb = volume of CO2 The decrease in volume of flue gas by second bulb = volume of O2 The decrease in volume of flue gas by third bulb = volume of CO.

3.15 REVIEW QUESTIONS 3.15.1 Fill in the Blanks 1. _______ is the C.G.S unit of heat. [Ans.: Calorie] 2. 1 K Cal = 3.968 B.T.H = _______ C.H.U. [Ans.: 2.2] 3. _______ is used for determining the calorific value of solid and liquid fuels. [Ans.: Bomb calorie meter] 4. Latent heat of steam is _______. [Ans.: 588 cal/gm] 5. The lightest temperature obtained with the fuel is known as _______. [Ans.: Pyrometric effect] 6. KJeldahl method is used for determination of _______. [Ans.: Nitrogen] 7. Aromatic hydrocarbons have _______ anti knocking properties than paraffins and olefins. [Ans.: Higher] 8. In flue gas analysis by orsat’s method _______ containing bulb is absorbed by CO2. [Ans.: KOH reduction]

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3.15.2 Multiple-choice Questions 1. Which of the following fuels possesses the maximum calorific value? (a) C = 84%, H = 6%, S = 4% and O = 6% (b) C = 84%, H = 12%, S = 1% and O = 1% (c) C = 90%, H = 5%, S = 2% and O = 3% (d) C = 95%, H = 2%, S = 1% and O = 2% [Ans.: b] 2. A good fuel should posses (a) High ignition temperature (c) High calorific value

(b) Moderate ignition temperature (d) Both b and c

[Ans.: d] 3. Ignition temperature of a fuel is the (a) Temperature at which the fuel can be stored safely (b) Lowest temperature at which the fuel must be preheated so that it starts burning (c) Temperature attained with the fuel is burnt (d) Temperature at which the fuel ignites for a moment, but doesn’t burn after then [Ans.: b] 4. Which of the following is not an advantage of gaseous fuels over solid and liquid fuels (a) They can easily be conveyed through pipelines to the actual place of use (b) They can be lighted at moments notice (c) They cannot be preheated by the heat of the hot waste gases (d) Their combustion can readily be controlled [Ans.: c] 5. Which of the following statements is true (a) Coke possesses better strength than coal (b) Coke burns with a long flame (c) Coke burns with a short flame (d) Sulphur content of coke is higher than that of coal from which it is obtained [Ans.: c] 6. Which of the following fuel gases possesses the highest calorific value (a) Water gas (b) Coal gas (c) Producer gas (d) Natural gas [Ans.: b] 7. Petrochemicals can be used to prepare: (a) PVC plastics (c) Terylene fibers [Ans.: d]

(b) Polystyrene plastics (d) None of these

8. The maximum temperature reached, when the coal is completely burnt in the theoretical amount of air is called: (a) Fusion temperature (b) Calorific intensity (c) Ignition temperature (d) None of these [Ans.: b]

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3.34 Engineering Chemistry 9. The calorific value of a coal sample is higher, if its: (a) Moisture content is high (b) Ash content is high (c) Volatile matter is high (d) Fixed carbon is high [Ans.: d] 10. Which of the following in coal decreases its calorific value: (a) Carbon (b) Hydrogen (c) Oxygen (d) Sulphur [Ans.: c] 11. Which of the following is not a characteristic of progressive transformation of wood to coal during coalification? (a) Fixed carbon content increases (b) Moisture content decreases (c) Volatile matter increases (d) Oxygen content decreases [Ans.: c] 12. Peat is: (a) Soft brown coloured (c) Pitch black coal [Ans.: b]

(b) Brown jelly like mass (d) Last stage of coalification

13. Which of the following contain highest percentage volatile matter? (a) Peat (b) Lignite (c) Bituminous coal (d) Anthracite [Ans.: a] 14. In orsats apparatus, potassium hydroxide is used to absorb: (a) Oxygen (b) Carbon dioxide (c) Carbon monoxide (d) Sulphur dioxide [Ans.: b] 15. Orsats apparatus is used to obtain: (a) Specific heats of components (b) Molecular weights of components (c) Gravimetric analysis of a gas mixture (d) Volumetric analysis of flue gases [Ans.: d] 16. Higher calorific value of fuel assumes that is: (b) Contains H2O in vapour form (a) Contains H2O in liquid form (c) Ignores the presense of H2O (d) Contain H2O in liquid and vapour forms [Ans.: d] 17. Stoichiometric quantity of air is the quantity of air required for complete combustion of fuel with (a) some excess oxygen (b) non oxygen left unused (c) 50% excess air (d) 100% excess air [Ans.: b]

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18. Analysis of flue gages is done by: (a) Boy’s gas calorimeter (c) Retort [Ans.: b]

Fuels and Combustion

3.35

(b) Orsat apparatus (d) Bomb calorimeter

19. Bomb calorimeter is used for determining the calorific value of (a) Solid fuel (b) liquid fuel (c) gaseous fuel (d) both a and b [Ans.: d] 20. Proximate analysis of fuel is determination of percentage of (a) C, H, N, S, H2O (b) C, H2O, ash, volatile matter (c) C only (d) useful heat evolved [Ans.: b] 21. Ultimate analysis of fuel is determination of percentage of (a) C, H, N, S, H2O (b) C, H2O, ash, volatile matter (c) sulphur only (d) fixed carbon only [Ans.: a] 22. Bomb calorimeter is used to determine: (a) HCV at constant pressure (c) HCV at constant volume [Ans.: c]

(b) LCV at constant pressure (d) LCV at constant volume

23. Incomplete combustion can be best judged by (a) smoky chimney exit (b) excess air in flue gases (c) measuring CO in flue gases (d) measuring O2 in flue gases [Ans.: c] 24. Gas with least calorific value is: (a) coal gas (c) producer gas [Ans.: c]

(b) water gas (d) natural gas

25. Main constituent of natural gas is (a) carbon monoxide (c) Hydrogen [Ans.: b]

(b) methane (d) ethane

26. The process of splitting bigger hydrocarbons into smaller hydrocarbons is called (a) pyrolysis (b) thermal decomposition (c) cracking (d) combustion [Ans.: c] 27. Iso-octane (2,2,4-trimethy pentane) has an octane rating of: (a) 100 (b) Zero (c) 50 (d) above 100 [Ans.: a]

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3.36 Engineering Chemistry 28. Which of the following possess zero octane number: (a) Iso-octane (b) petrol (c) n-heptane (d) LPG [Ans.: c] 29. Suitability of a diesel fuel is determined by: (a) Octane rating (b) percentage of carbon (c) length of hydrocarbon chain (d) Cetane number [Ans.: d] 30. For good performance, the hydrocarbon molecules in a diesel fuel should be (a) Branch-chained (b) side-chained (c) straight–chained (d) aromatic [Ans.: c] 31. The cetane rating of hexadecane is: (a) 100 (c) 50 [Ans.: a]

(b) 0 (d) none of these

32. Which of the following is used as a jet engine fuel: (a) LPG (b) Kerosene (c) Power alcohol (d) coal [Ans.: b] 33. Main constituent of LPG is (a) Methane (c) Benzene [Ans.: d]

(b) Propane (d) Butane

34. Alcohol has an octane number of about (a) 50 (c) 90 [Ans.: c]

(b) 60–70 (d) 25

35. Alcohol-blended petrol possesses (a) Better calorific value (c) Poorer-antiknock properties [Ans.: b]

(b) Better anti knock properties (d) None of these

36. In Bergius process of preparing synthetic petrol by (a) Passing water gas over heated powdered coke under pressure (b) Catalytic hydrogenation of coal (c) Heating coal along under pressure (d) Cracking of heavy oil [Ans.: b]

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37. A good fuel should possess: (a) Low calorific value (c) High calorific value [Ans.: c]

Fuels and Combustion

3.37

(b) High ignition temperature (d) High ash content

38. Anthracite: (a) is lowest rank coal (b) contains high percentage of carbon (c) contains high percentage of volatile matter (d) high calorific value and high carbon percentage [Ans.: d] 39. An example of primary fuel is (a) Natural gas (c) Wood charcoal [Ans.: a]

(b) Petrol (d) Coke

40. Lignite is: (a) Lowest rank coal (c) Used in metallurgy of iron [Ans.: a]

(b) Highest rank coal (d) Contains no moisture

3.15.3 Short Answer Questions 1. Define fuel and give some examples. Ans.: Fuel is a combustible substance containing carbon as the major constituent which on proper burning gives large amount of heat that can be used economically for domestic and industrial purposes. Examples are coal, petrol, diesel, etc. 2. Give classification of fuels according to the occurrence. Ans.: According to the occurrence, the fuels are classified into natural (primary) and secondary (derived) fuels. 3. What are the units of heat and their inter-conversions? Ans.: Units of heat are calorie, kilo calorie, British thermal unit and centigrade heat unit. 1 kcal = 1000 cal = 3.968 BTU = 2.2 CHU 4. Define calorific value and give relation between higher and lower calorific values. Ans.: Calorific value is the total quantity of heat liberated by the complete combustion of one unit mass/volume of a fuel in oxygen. LCV = HCV – latent heat of water vapour formed 5. What kinds of calorimeters are used for determining calorific value of solid, liquid and gaseous fuels? Ans.: Bomb calorimeter is used for determining the calorific value of solid and liquid fuels. Junker’s calorimeter is used for determining the calorific value of gaseous fuels.

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3.38 Engineering Chemistry 6. What are the characteristics of a good fuel? Ans.: High calorific value, moderate ignition temperature, low moisture content, low non-combustible matter content, etc. 7. Name some of the petroleum products. Ans.: Liquified petroleum gas, gasoline, petrol, kerosene, diesel oil, heavy oil, etc. 8. What are the ovens used for preparation of metallurgical coke? Ans.: Beehive oven and Otto Hoffman’s by-product oven. 9. Explain cracking with a suitable example. Ans.: The process of breakdown of high molecular weight hydrocarbons of high boiling points into simple, lower molecular weight hydrocarbons of low boiling points is known as cracking. Cracking → C5 H12 + C5 H10 Ex: C10 H 22 

Decane

n-pentane Pentane

10. What is meant by flue gas? Ans.: The mixture of gases like CO2, CO and O2 exhaust of the combustion chamber is called flue gas. 11. What is the importance of analysis of flue gas? Ans.: The analysis of flue gas either from a furnace or from an engine’s exhaust would give an idea about the efficiency of the combustion process. If the flue gas contains considerable amount of CO, it indicates that incomplete combustion is occurring and it also indicates the short supply of O2 for combustion, and this will lead to wastage of fuel. 12. What happens if the flue gas contains considerable amount of O2? Ans.: It indicates that the O2 supply is very much in excess, and it results in loss of heat. 13. Which apparatus is used in the analysis of flue gas? Ans.: Orsat’s apparatus. 14. In Orsat’s apparatus, which gases are absorbed by which solutions? Ans.: Potassium hydroxide solution – only CO2 Alkaline pyrogallic acid – CO2 and O2 Ammonical cuprous chloride – CO, O2 and CO2

3.15.4 Descriptive Questions Q.1 a. W hat do you understand by the term knocking in IC engines? Explain its relationship with chemical constituents of fuels. b. A sample of coal contains 60% carbon, 33% oxygen, 6.0% hydrogen, 0.5% sulphur, 0.2% nitrogen and 0.3% ash. Calculate GCV and NCV of coal. Q.2 Distinguish between the followings: a. Gross and net calorific values b. Octane number and centane number

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3.39

Q.3 a. How coal is graded? Explain its calorific value of coal. b. Give the advantages and disadvantages of coal over gaseous fuels. Q.4 a. Explain the proximate analysis of coal and its significance. b. Distinguish between low-temperature carbonisation and high-temperature carbonisation. Q.5 a. W hat is metallurgical coke? How it is superior than coal? Explain the manufacture of the metallurgical coke by Otto Hoffman’s by-product coke oven method. List the various by-products obtained. b. Define octane number of gasoline. Why is ethylene dibromide added, when tetra ethyl lead is used as an anti-knock? Q.6 A fuel containing 92% C and 4% H2 by mass was burnt in 90% of air of that required for complete combustion. Find out the % of composition of dry product of combustion by mass of H 2 is burnt completely and no carbon is left behind. Q.7 Give brief note on the following: a. Explain how fuels are classified with suitable examples. b. Give the comparison between solid, liquid and gaseous fuels. c. What are the characteristics of a good fuel? Q.8 Explain the significance of the following constituents present in coal: a. Moisture b. Volatile matter c. Ash d. Fixed carbon Q.9 a. Discuss the relative merits and demerits of solid, liquid and gaseous fuels. b. Explain the significance of the following constituents present in coal. Q.10 a. How a calorific value of a gaseous fuel is determined by Junker’s gas calorimeter? Describe the experiment with a neat diagram. b. Calculate gross and net calorific value of a gaseous fuel from the following data. Vol. of gaseous fuel burnt at STP -0.09 m3, weight of water used for cooling 25.0 kg, temperature of inlet water 25 °C, temperature of the outlet water 30.0 °C, weight of water produced by steam condensation 0.02 kg latent heat of steam 587 kcal/kg. Q.11 a. What are the constituents of petroleum? Describe the origin of petroleum. b. Give an account of production of petrol from crude oil. Q.12 The analysis of a gas has the following composition: H2 = 14%, CH4 = 2%, CO = 22%, CO2=5%, N2 = 55%, O2 = 2%. Find the air required for the combustion of 1 m3 of the gas. If 45% excess air is supplied, find the volume analysis of the products. Q.13 a. Define a fuel? How chemical fuels are classified and give examples for each? b. What is meant by calorific value of a fuel? Define calorie and kilocalorie.

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3.40 Engineering Chemistry Q.14 Explain the significance of the following constituents present in coal: a. Total carbon b. Hydrogen c. Nitrogen Q.15 a. What are chemical fuels? Give complete classification of chemical fuels with examples. b. What are different types of fuels? What are the characteristics of a good fuel? c. Mention the criteria for selecting good fuel. d. Distinguish between solid, liquid and gaseous fuels. Q.16 a. What is meant by calorific values of a fuel? b. Describe how the calorific value of a solid fuel is determined using a bomb calorimeter. c. What are the fuels used for determination of water equivalent of bomb calorimeter and why? Q.17 a. Differentiate proximate and ultimate analysis of coal. b. Discuss the importance of ultimate analysis of coal. Q.18 a. What a good fuel must have low ash content? Or what is role of ash on coal? b. How is nitrogen determined in a solid fuel? c. What is the significance of a volatile matter in coal? d. How is ranking of coal make based on ultimate analysis? Q.19 a. What are the advantages of liquid fuels over solid fuels? b. Differentiate between coal and coke. c. Explain carbonisation of coal. d. Why is coke preferred to coal in metallurgical purposes? e. Why are gaseous fuels more advantageous than solid fuels? Q.20 a. Write short note on Beehive coke oven. b. Why is peat not considered as an economical fuel? Q.21 What is crude oil? Write short note on refining of crude petroleum. What are the various fractions obtained from petroleum? Mention the industrial uses to which they are put. Q.22 a. W hat are the structural features of hydrocarbons in unlead petrol and diesel? What are the structural factors that promote its high value? b. W hat is the significance of octane number and cetane number and for which these are used? How these can be improved? c. W hy is C2H4Br2 added, when TEL is used as an anti-knock? d. W hat types of compounds nowadays are being added to petrol to improve octane rating? Q.23 a. W hat is meant by cracking of petroleum? Explain? Fluidised-bed catalytic method of obtaining gasoline. Give its mechanism. b. W hat are the advantages of catalytic cracking process? Describe, with a neat diagram, the fixed-bed catalytic cracking process.

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3.41

c. D ifferentiate between thermal and catalytic cracking. d. W hat are the advantages of catalytic cracking over thermal cracking? e. W hat is meant by knocking? How is it related to chemical constitution? Describe the functions of TEL. Explain octane number and cetane number. Q.24 a. What is LPG? What are the advantages of LPG over gaseous fuels? b. Write the approximate compositions and calorific values of water gas and producer gas. Q.25 a. Define flue gas. How the analysis of flue gas is done by the Orsat apparatus? What conclusion can be drawn from the experiment? What is the significance of this analysis? b. H ow distinction can be made between complete and incomplete combustion of fuel? c. W hat is leaded petrol? Q.26 Write short notes on the following: a. Catalytic converter b. F lue analysis and its significance Q.27 a. What is the principle of bomb colorimeter? b. How gross calorific value of a solid fuel determined by Bomb Calorimeter? Write Dulong’s formula for calculating calorific value of fuel from its ultimate combustion data. c. D iscuss Beehives oven method for the manufacture of coke. Q.28 a. Describe the fractional distillation of petroleum. b. W hat do you understand with the knocking of fuel? Report the ways to improve the antiknocking characteristic of a fuel. Q.29 Write short note on metallurgical coke. Q.30 A coal sample has the following composition: C = 90%; H = 3.5%; O = 3%, S = 0.5%; and N2=1%; the remaining being ash. Calculate the theoretical volume of air required at 27 °C and 1 atm pressure when 100 kg of the coal is burnt. Q.31 The composition by weight of a coal sample is C = 81%; H = 5%; O = 8.5%; S = 1%; N = 1.5%; Ash = 3%: a. Calculate the amount of air required for the complete combustion of 1 kg of the coal. b. Calculate the gross and net calorific values of the coal sample. Given that the calorific values of C, H and S are 8,060 kcal/kg; 3,400 kcal/kg and 2,200 kcal/kg, respectively. Q.32 A producer gas has the following composition by volume: CO = 30%; H 2 = 12%; CO2 = 4%; CH4 = 2% and N2 = 52%. When 100 m3 of the gas is burnt with 50% excess air used, what will be the composition of the dry flue gases obtained? Q.33 A hydrocarbon fuel on combustion gave the flue gas having the following composition by volume: CO2 = 13%, O2 = 6.5% and N2 = 80.5%. Calculate (a) the composition of fuel by weight and (b) the percentage of excess air used. Q.34 Calculate the approximate calorific value of a coal sample having the following ultimate analysis: C = 80%; H = 3.5%; S = 2.8%; O = 5.0%; N = 1.5% and ash = 7.2%.

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ALTERNATE ENERGY RESOURCES Energy conservation is the foundation of energy independence

4.1

INTRODUCTION

Every day, a lot of carbon dioxide is being released due to different developmental activities that accumulates around the atmosphere. Such release of carbon dioxide is one of the causes of global warming. One answer to reduce the global warming effect is to replace and retrofit current traditional energy resources with alternative energy sources which perform comparably better without emitting carbon dioxide. At the same time, with increasing demand for energy and with fast depleting conventional sources of energy such as coal, petroleum, natural gas and so on, the non-conventional sources of energy such as energy the sun, wind, biomass, tidal energy, geo-thermal energy and even energy from waste material are gaining importance. This energy is abundant, renewable, pollution-free and eco-friendly. The importance of renewable energy was recognised in the country in the early 1970s. The renewable energy programme started with the establishment of the Department of Non-conventional Energy Sources (DNES) in 1982. The Indian Renewable Energy Development Agency (IREDA) was set up in 1987. In 1992, the DNES was converted into the Ministry of Non-conventional Energy Sources (MNES) which has taken several steps to create a suitable atmosphere for harnessing non-conventional sources of energy. Today, India has one of the largest programmes for renewable energy. Depending on their usage, energy resources are broadly divided into two categories—conventional and nonconventional energy resources.

4.1.1

Conventional or Traditional Energy Resources

The energy resource that has been used from ancient period is known as conventional or traditional energy resources. A majority of these energy resources are non-renewable. For example, coal, fuel wood, crude oil, natural gas, etc. All conventional sources will become rare, endangered and extinct, as they produce lots of carbon dioxide that adds to the greenhouse effect in the atmosphere.

4.1.2

Nonconventional Energy Resources or Renewable Energy Sources

The energy sources which will not get exhausted even after continuous and excessive use and are abundantly available, continuously replenished by nature are nonconventional energy resources. These energy resources are environmentally and ecologically safe. For example, solar energy, wind energy, tidal energy, biogas, geothermal energy, nuclear energy, etc.

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4.2 Engineering Chemistry

4.1.3 Alternative Energy Alternative energy is commonly referred to as any energy source that is an alternative to fossil fuel. According to previous reports, coal as an alternative to wood, petroleum as an alternative to whale oil, alcohol as an alternative to fossil fuels, coal gasification as an alternative to petroleum etc., were common; however, this is completely different. Depending on availability, demand and sustainability, the energy resources which are used as alternative energy source has changed considerably from time to time. Alternative energy plays vital role to meet the future demands of energy and environment; hence, advance research in this area started in a multidisciplinary way. Let us examine the various definitions of alternative energy. Oxford Definition Energy fuelled into ways that does not use up natural resources or harm the environment. Princeton WordNet Energy derived from sources that does not use up natural resources or harm the nature. Natural Resource Defence Council Energy that is not popularly used and is usually environmental sound such as solar or wind energy. Material and Management Fuel source that are other than those derived from fossil fuels typically used inter changeable for renewable energy. For example, wind energy, solar energy, biomass, wave and tidal energy. Torridge District Council Energy generated from alternatives to fossil fuel, but need not be renewable. Climate Change 2007 Energy derived from nontraditional sources. For example, compressed natural gas, solar energy, hydroelectric and wind energy. Hence, common alternative energy resources are solar energy, water energy (hydroelectric power and tidal power), wind energy, geothermal energy, biomass energy, nuclear energy, etc.

4.2 NON-CONVENTIONAL ENERGY SOURCES AND STORAGE DEVICES 4.2.1 Solar Energy The sun is a source of enormous energy. Solar energy originates with thermo-nuclear fusion in the sun, involving the fusion of hydrogen into helium either through the proton-proton chain or through the carbon-nitrogen cycle. The Earth receives solar energy as a radiant energy which reaches the top of the atmosphere at 1,366 watts per square meter. This energy ranges from ultraviolet, visible and infrared light. About half of this energy actually makes it to the Earth’s surface, 30% is reflected and

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Alternate Energy Resources

4.3

20% is absorbed by the atmosphere. Thus, when the sun is directly overhead, full sunlight can deliver about 700 watt per square meter and at that rate, the sun can deliver 700 MW of power to an area of 390 square miles. The total amount of solar energy reaching the earth is vast and almost beyond belief. For an example, one year’s expenditure of fossil fuel in the USA is equivalent to just 40 minutes of sunlight striking the particular land surface of the USA. The solar energy technology broadly comprises thermal conversion and photo conversion. Thermal conversion takes place through direct heating, ocean waves and currents and wind. Photo conversion includes photosynthesis, photochemistry, photo electrochemistry, photo galvanism and photovoltaics. Solar radiation is collected and converted by natural collectors such as the atmosphere, the ocean and plant life, as well as by man-made collectors of many kinds (Figure 4.1). There are a number of solar technologies by which it can be harnessed (Figure 4.2). Solar radiation

Atmosphere

Ocean

Structures

Wind

OTEC∗

Passive

Man-made collectors

Heat

Quantum process

Plants

Biomass

Useful energy ∗OTEC = Ocean thermal energy conversion

Figure 4.1 Natural and man-made collectors Process heat biomass Solar thermal heating & cooling

SUN

Solar radiation

Heat

Photo voltaics Solar thermal OTEC windmills

Electricity

Windmills

Mechanical energy

Biochemical conversion

Chemical/fuels

Figure 4.2 Solar technologies

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4.4 Engineering Chemistry The earliest use of direct solar energy by mankind was drying the body or warming it in the sun during winters. Indeed, drying of clothes, fodder, timber, agricultural and animal products, salt water (to get salt) and passive space heating remained the most extensive form of use of direct solar energy in the history of mankind. All other active solar technologies or devices for harnessing direct solar energy have fairly recent origins (Figure 4.2). A variety of active solar collectors provide a broad range of applications as follows: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii)

Solar heating of water Solar space heating of buildings Solar air-conditioning Solar refrigeration Solar drying Solar cooking Solar green-houses Solar furnaces Solar desalination Salt production Solar electricity—thermal Solar electricity—photovoltaic

Solar Heating of Water Solar water heating is very popular in warm, sunny climates. Flat plate collectors for heating water consists of a thin broad box with a glass or plastic top and a black bottom in which water tubes are embedded as shown in Figure 4.3. These collectors are faced towards the sun and the black bottom gets hot when it absorbs sunlight. Thus, water circulating tubes are heated and are conveyed to a tank where it is stored. Solar heating system may be active or passive. The heated water is moved by means of pumps in an active system. However, in a passive system, the collector is lower than the tank. By natural convection, the heated water rises into the tank from the collector and the cool water descends into the collector from the tank. This is shown in Figure 4.4. The tank will usually have a source of auxiliary heat (electric or gas) in order to get the temperature to a desired level or to provide heat when solar energy is insufficient. Solar Space Heating of Buildings Flat-plate collectors such as those used in water heating with same concept and less expensive option can be used for space heating as only air circulation through the collector box is necessary. In nonfreezing climates, simple water convection systems may suffice. In freezing climates, an antifreeze fluid is circulated. Solar heat is augmented by an auxiliary heat source in the hot-water tank. The efficiency of solar space heating can be enhanced by proper designing of a building to act as its own collector. Here, the basic principle is to have windows facing the sun. On cool days, it depends on sun’s angle of incidence and sunlight can heat the interior of the building; however, at nights, insulated drapes or shades can be pulled down to trap the heat inside. Excessive heat in hot days can be avoided by using an awning or overhang to shield the window from the sun.

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Alternate Energy Resources

Sunlight

4.5

Tubes through which water circulates

Heat

Black surface

Insulation

Glass or plastic window

Figure 4.3 Principle of a flat-plate solar collector

Element

t-w

ate

Hot-water tank

Hot-water out

Water circulates by convection

Auxiliary heat (electric or gas)

Collector

Cold water in

Figure 4.4 Solar water heaters Solar Air-conditioning Solar air-conditioning includes solar-powered refrigeration system of ranking cycle system, absorption refrigerator systems and solar-regenerated desiccant cooling systems. In such type of system, ambient air is adiabatically cooled, dehumidified, cooled both sensibly and evaporatively and then ducted to the living area. In the regenerative stage, air is evaporatively cooled, heated as it cools the supply air stream, heated again by solar collectors and humidified.

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4.6 Engineering Chemistry Solar Refrigeration Solar refrigeration is closely related to air-conditioning. It is generally required for food preservation or for storage of medical and biological materials. Although most of the units/machines fabricated are simple in design, they are generally too complicated to operate and, therefore, are not usable by the people. Solar Drying Solar drying of agriculture and animal products is the most ancient, traditional and wide-spread method of utilising direct solar energy. Agricultural products such as grain, hay, copra, fruits, nuts and vegetables are still-dried with sun drying all over the world, including highly industrialised and well-developed countries. This method is followed because it is the cheapest and the simplest way to dry crops and abundant availability of sunlight is also ensured. Solar Cooking A solar cooker is a perfectly insulated shallow rectangular metal box, and inner side of this box is blackened and fitted with a flat glass cover. When this cooker is placed in sunlight, the solar energy penetrates the glass cover and is absorbed by the inner blackened surface. Thus, temperature will increase inside the box, and the food gets cooked quickly. The collector area of such a solar cooker can be increased by providing a plane reflector mirror of size equal to the size of the box and is hinged on one side of the glass frame. With the help of reflector mirror, a temperature rise of 15°C–20°C can be achieved inside the solar cooker. Solar Greenhouses A greenhouse is a closed structure covered with transparent material (glass or plastic) which acts as a solar collector and utilises solar radiant energy for the growth of plants. The incoming shortwave solar radiations can pass through the green house glazing; but the longwave thermal radiations emitted by the objects within the greenhouse cannot escape through the glazed surface. As a result, the radiations get trapped within the greenhouse and result in an increase in temperature. Further, the air inside the greenhouse gets enriched with carbon dioxide and the moisture loss is reduced due to restricted transpiration. All these factors help the plant growth to sustain during night and colder months. Solar Furnaces Extremely high temperatures with very clean condition of around 3,500°C can provide solar furnaces. This can be used to melt refractory material. In a solar furnace, high temperature is obtained by concentrating the solar radiations on to a specimen using a number of heliostats (turnable mirrors) arranged on a sloping surface. The biggest advantage of a solar furnace is that heating can be accomplished without any contamination and temperature can be easily controlled by changing the position of the material in focus. It is anticipated that in future, solar furnaces can be utilised in the production of nitric acid and fertilisers from air. Solar Desalination In this method, solar radiation is admitted through a transparent air-tight cover of sloping sheets of glass into a shallow blackened pool containing brine. The water evaporates from the brine when solar radiations pass through the cover. The vapours produced get condensed as purified water on the cooling

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Alternate Energy Resources

4.7

Solar radiation Potable distilled water trough (leads to storage tank)

Brine inlet

Brine

Excess brine to waste

Figure 4.5 Basin type solar-still glass cover, flow down the sloping roof, collected into a water storage tank (as shown in Figure 4.5). The excess brine that has not evaporated is run to waste. The cost of distilled water per litre obtained by this process is cheaper than distilled water obtained by other electrical energy-based processes. Salt Production It is the most widely used method for salt production in many developing countries of the world. The basic concept is that in areas where evaporation exceeds rainfall, a shallow pool of brine is exposed which results in evaporation of water, leaving behind the salt. Solar Electricity—Thermal Solar energy may be used to heat a fluid, which then generates electricity through a conventional heat engine. To obtain an adequate working temperature, some form of concentration of solar energy is required, so that there is little contribution from diffused sunlight for most designs. Broadly, the systems fall into two categories: (i) Systems in which individual mirrors track the sun continuously (ii) Systems in which mirrors are fixed In continuous tracking systems, a large number of plane or curved mirrors is used. Each mirror is steered to reflect sunlight onto a single tower mounted boiler and gives a high temperature with high efficiency. However, mirrors require complex, rugged and accurate mechanisms. Non-tracking systems consist of assemblies of trough-shaped collector, aligned east to west. Above each collector, the absorber is fitted in the form of a tube. A general plan of a solar-thermal electric power plant is shown in Figure 4.6. It is expected that cost-effective and utility scale solar-thermal power plants might be built in the near future. Solar Electricity—Photovoltaic A solar cell—more properly called a photovoltaic, or a PV cell—directly converts incident solar radiation to electrical current. A PV cell looks like a simple wafer of material with one wire attached to the top and one to the bottom. As sunlight shines on the wafer, it puts out an amount of electric current roughly equivalent to that emitted by a flashlight battery. Thus, PV cells collect light and convert it to electric power in one step. However, several cells can be connected together to obtain large amount of power.

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4.8 Engineering Chemistry

Thermal energy storage

Receiver

Solar thermal collector

Heat exchanger Steam for process

Feed water

Electrical power plant

Electricity

Figure 4.6 General plan of a solar-thermal electric power plant Working Principle PV cells consist of two very thin layers of semiconductor materials separated by a junction layer. The lower layer has atoms with single electrons in their outer orbit which are easily less. The upper layer has atoms lacking electrons in their outer orbit; these atoms readily gain electrons. When the kinetic energy of light photons striking the two-layer “sandwich” dislodges electrons from the lower layer, it creates a current that can flow through a motor or some other electrical device back to the upper side. The major material used in PV cells is silicon. PV cells consist of a single crystal of p-type silicon with a surface layer of n-type silicon. When light falls on the junction, the electrons and holes move in opposite directions across the p-n junction and electric current will flow if an external circuit is connected. This is shown in Figure 4.7. Cost The cost of PV power (cents per kilowatt-hour) is the cost of the PV cells divided by the total amount of power they may be expected to produce over their lifetime currently around 25 cents per kilowatthour while the cost of power from other power alternative is 6–12 cents per kilowatt-hour for residential electricity. PV power had its first significant application in the 1950s in the solar panels of space satellites. The cost comes down dramatically if more efficient cells with less expensive production techniques are evolved. Uses The uses of PV cells are as follows: (i) PV cells are widely used in watches, calculators, toys, etc.

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Alternate Energy Resources

4.9

(ii) T he panels of PV cells provide power for traffic signals, radio transmitters, light houses, irrigation pumps, earth orbiting satellites, etc. p-n Junction A

Radiation

n Current flow

Electron flow

Figure 4.7 Semiconductors functioning as solar cells Future of Solar Energy Solar electricity is growing at a phenomenal rate due to adverse environmental effects and logging of non-renewable energy resources. The sun provides power only during the day, but 70% of electrical demand occurs during the day for industries, offices and stores that are in operation. Thus, remarkable savings can be achieved by using solar panels for daytime requirements and continuing to rely on conventional sources at night.

4.2.2 Wind Energy The research pertaining to wind energy dates back several decade to the 1970s when the NASA developed an analytical model to predict wind turbine power generation during high winds. Today, both Sandia laboratory and national renewable energy have programmes dedicated to wind research. The Field Laboratory Operator is Optimised Wind Energy (FLOWE) at Caltech was established to research alternative approaches to wind energy. The sun is the origin of wind energy. Wind is a form of solar energy and is caused by the uneven heating of the atmosphere by the sun, the irregularities of the earth surface and rotation of the Earth. Wind flow patterns are modified by the Earth’s terrain bodies of water and vegetation cover. Kinetic energy in the wind can be used to run wind turbines but the output power depends upon the wind speed. Turbines generally require a wind in the range of 20 km/h. Wind turbine converts kinetic energy in the wind into mechanical power. This mechanical power can be used for a specific task or the generator converts mechanical power into electricity to power homes, businesses, schools and so on.

Figure 4.8 Wind turbines

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4.10 Engineering Chemistry Like air draft propeller blades, wind turbines (Figure 4.8) turn in the moving air and power an electric generator that supplies an electric current. Simply stated, a wind turbine is the opposite of a fan. Instead of using electricity to make wind, it is using wind to make electricity. The wind turns the blades, which spin a shaft, which connects to a generator and makes electricity. Wind energy is a free, renewable resource; the wind turbines set-up is a bit expensive as well.

4.2.3 Geothermal Energy Geothermal energy is produced by tapping into the thermal energy created and stored within the Earth’s crust (Figure 4.9). It is considered sustainable because thermal energy is constantly replenished. Several entities such as the National Renewable Energy Laboratory and Sandia National Laboratories are conducting research towards the goal of establishing a proven science around the geothermal energy. Geothermal energy comes from the heat within the earth. The word geothermal comes from the Greek words geo meaning earth and thermal meaning heat. Geothermal energy has been used for thousands of years in some countries for cooking and heating. People around the world use geothermal energy to produce electricity to heat buildings, green houses and for other purposes. Power plant

Turbine

Generator

Load

Well

Heat exchanger Geothermal reservoir

Magma

Production Well

Figure 4.9 Geothermal plant

Injection Well

There are four kinds of geothermal sources—hydrothermal, geo pressured, hot dry rock and magma.

4.2.4 Water Power The energy of falling or fast-running water can be harnessed and used in the form of motive energy or temperature difference. Since water is about a thousand times heavier than air, even a slow flowing stream of water can yield great amount of energy. Hydroelectric Energy Hydroelectric energy is a term usually referred to in the context of hydroelectric dams. Hydroelectric power stations capture the kinetic energy of moving water and give mechanical energy to turbines. The moving turbines then convert mechanical energy into electrical energy through generators.

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Alternate Energy Resources

4.11

This energy has been exploited for centuries. Farmers since the ancient Greeks have used water wheels to grind wheat into flour. Placed in a river, a water wheel picks up flowing water in buckets located around the wheel. The kinetic energy of the flowing river turns the wheel and is converted into mechanical energy that runs the mill (Figure 4.10). In the late 19th century, hydropower became a source for generating electricity. The first hydroelectric power plant was built at Niagara Falls in 1879. In 1881, street lamps in Canada were powered by hydropower. In 1882, the world’s first hydroelectric power plant began operating in Appleton, Wisconsin. A typical hydro plant is a system with three parts—an electric plant where the electricity is produced, a dam that can be opened or closed to control water flow and a reservoir where water can be stored. The water behind the dam flows through an intake and pushes against blades in a turbine, causing them to turn. The turbine spins a generator to produce electricity. The amount of electricity that can be generated depends on how far the water drops and how much water moves through the system. The electricity can be transported over long-distance electric lines to homes, factories and businesses. Generation of electricity from a small-sized hydropower source is a low-cost, environment friendly and renewable source of energy. Small and mini hydel projects have the potential to provide energy in remote and hilly areas where extension of grid system is uneconomical.

Sluice gates

High pressure water

Water storage reservior

Generator

Dam

Figure 4.10 Hydroelectric power plant Tidal Energy Tidal energy is a form ofhydropowerthat converts the energy obtained fromtidesinto useful forms of power, mainly electricity. Sea water keeps rising and falling alternatively twice a day under the influence of gravitational pull of moon and sun. This phenomenon is known as tides. Tidal power refers to capturing energy from the tides in horizontal direction. Tides come in, raise water levelsin a basin and tides roll out. The water is made to pass through turbine to get out of the basin. Therefore, power generation through this method has a varying degree of success (Figure 4.11). There are two types of tidal energy systems that can be used to extract energy—kinetic energy, the moving water of rivers, tides and open ocean currents; andpotential energyfrom the difference in height between high and low tides. The first method—generating energy from tidal currents—is becoming more popular because people believe that it does not harm the environment as much as barrages ordams. Many coastal sites worldwide are being examined for their suitability to produce tidal energy.

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4.12 Engineering Chemistry

Tidal power

Figure 4.11 Tidal energy

4.2.5 Biomass Biomass refers to all plant material and animal excreta when considered as an energy source. Some important kinds of biomass are inferior wood, urban waste, farm animal and human waste. Biomass as an energy source can be either used directly via combustion to produce heat or indirectly after converting it into various forms of biofuels. Conversion of biomass to biofuels can be achieved by different methods such as thermal, chemical and bio chemical methods. Wood remains the largest biomass energy source. Since biomass includes plant or animal matter that can be converted into fibers or other industrial chemicals including biofuels, industrial biomass can be grown from numerous types of plants. Biomass, biofuel and bio gas are burned to produce heat/power and consequently, sulphur oxide, nitrous oxide and particular matter that pollute the environment are produced from this combustion. Biogas Biogas is the gas resulting from an anaerobic digestion process (Figure 4.12). A biogas plant can convert animal manure, green plants, waste from agro industry and slaughterhouses, paper production, sugarcane production, sewage and so forth into combustible gas. Biogas is a mixture of methane, carbon dioxide, water and hydrogen. Biogas power generation flow chart Oxygen 2O2

Organic waste

Anaerobic digester

Methane CH4

Power generator

Electric power

Heat

Other substances

Carbon dioxide and Water CO2 + 2H2O

Figure 4.12 Flow chart of biogas power generation

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Biofuel Biofuel is any fuel that derives from biomass; this may be recently living organisms or their metabolic byproducts, such as manure from cows, buffaloes etc., Typically, biofuel is burnt to release its stored chemical energy. Biomass can be directly used as fuel or to produce liquid biofuel. Agriculturally produced biomass fuels such as biodiesel, ethanol and bagasse (byproduct of sugarcane cultivation) can be burnt in internal combustion engines or boilers. Algae Fuel Algae fuel is a biofuel derived from algae. During photosynthesis, algae and other photosynthetic organisms capture carbon dioxide and sunlight and convert it into oxygen and biomass. This is usually done by placing the algae between two panes of glass. The algae create three forms of energy fuel— from its growth cycle, heat can be generate, the oil derived from the algae is used as biofuel and on maturity, algae can give biomass. The heat can be used to power building systems such as heating of water or to produce energy. Biofuel is oil extracted from algae upon maturity and is used to create energy similar to the use ofbiodiesel. Biomass is the matter left over after extracting oil and water and can be harvested to produce combustible methane for energy production. Biomass Briquettes Biomass briquettesare developed as an alternative to charcoal. Almost any kind of plant matter compressed into briquettes has about 70% the calorific value of charcoal.

4.2.6 Nuclear Energy With growing realisation that the supply of fossil fuels is depleting fast and is very limited, nuclear energy has gained importance and has become essential as an alternative source of energy. Due to rapid increase in industry and urbanisation, along with booming increase in population, it is estimated that the present fossil fuels (coal and petroleum) will not last more than a few decades to cater to the demand for electric power. Therefore, attention has been focused mainly on nuclear energy, which seems to offer an infinite source of energy. This field began with the discovery of radioactivity. The release of tremendous amount of energy through nuclear reaction became possible with the development of artificial radioactivity. Nuclear energy is the energy released when the nuclei of certain atoms undergo induced reactions such as fission and fusion. The materials that make such energy available are called nuclear fuel. Nuclear fuels release energy by an entirely different mechanism as compared to chemical reactions. In nuclear fuel, different elements are produced, and some of the binding energy of the nucleus is released, during bombardment of neutron on radioactive element such as uranium-238, etc. Therefore, a large amount of energy is released in less than a millionth of a second. For example, complete fission of one kg of uranium provides the energy equivalent of about 2 × 107 kWh; such amount of energy can be obtained by burning of about 3,000 tonnes of high grade coal. Mass Defects and Nuclear Binding Energy The mass of an atomic nucleus has been found to be always less than the sum of the masses of the constituent nucleons (i.e., protons, neutrons and electrons). This difference is called mass defect. The mass thus lost appears in the form of energy in accordance with Einstein’s mass-energy relationship; the energy emitted is called the binding energy of the nucleus.

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4.14 Engineering Chemistry Consider an isotope having atomic number ‘Z’ and mass number ‘A’. Evidently, its atom contains protons = Z, electrons = Z, Neutrons = (A − Z). Let mp, me and mn respectively represent the masses of proton, electron and neutron. Thus, the calculated mass of this isotope is M′ = Zm p + Zme + ( A − Z )mn ∴

[

= ZmH + ( A − Z)mn

mp + me = mH mass of hydrogen atom]

Let M = actual mass of the element Mass defect = M ′ − Μ = ∆M ∆M = ZmH + (A − Z)mn − M Thus, the mass defect (∆M) is the loss of mass in the formation of nucleus from its constituents. This loss of mass represents the amount of energy, which would be released if an atom of mass number A were synthesised from its constituents. This form of energy is in accordance with Einstein’s massenergy relationship. Thus, the binding energy of the nucleus may be defined as the energy released during the formation of a nucleus from its constituents nucleons. Thus, if ∆M is the mass defect, then, binding energy B.E. = ∆MC2, where C = Velocity of light (3 × 108m/s) B.E. = [ZmH + (A − Z) mn − M] C2 The binding energy per nucleon is calculated by dividing the binding energy of a nucleus by the number of nucleons. The binding energy corresponding to one atomic mass defect is 931.5 MeV (million electron volts) B.E. 1 Binding energy nucleon, = [ ZmH + ( A − Z)mn − M ]C2 A A The greater is B.E./nucleon, the greater is the stability of the nucleus. The energy equivalent to one atomic mass unit (1 amu) is given by, Mass of 1 atomic unit = 1.6604 × 10 −24 Velocity of light (C) ≈ 2.998 × 1010 cm/sec Thus E = (1.6604 × 10 −24 )(2.998 × 1010 )2 erg 1..6604 × 10 −24 (2.998 × 1010 )2 MeV 1.602 × 10 −6 = 931.5 MeV =

Example to Calculate the Binding Energy per Nucleon The formation of helium atom can be written as follows: 2 ⋅ 11 p + 2 ⋅ 01 n → 42 He Change in mass is given as follows: ∆M = 2m[ 11 p]+2m[ 01 n] − m[ 42 He] = 2m[ 11 H]+2m[ 01 n] − m[ 42 He] = 2(1.00782505) + 2(1.0086649) − 4.0026033 = 0.0303766 amu

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Alternate Energy Resources

4.15

Since the energy equivalent to 1 amu of mass is 931.5 MeV, the energy change during the formation 4 2 He of atom is ∆E = (0.0303766 amu)(931.5 MeV/amu) = 28.2956 MeV Thus binding energy B.E.= 28.2956 MeV Further, since there are four nucleons in 42 He nucleus, the binding energy per nucleon (EB/A) is obtained by dividing the total binding energy with the number of nucleons, that is, E B 28.2956 = = 7.07390 MeV A 4 The binding energy per nucleon is a direct measure of the stability of the nucleus. Binding Energies and Stability of Nuclei Mean binding energies, that is, the binding energies per nucleon, drastically increase with increasing atomic mass in lighter elements, reach a maximum at mass numbers 55−58, and gradually decrease in the case of heavier elements. The mass number 55–58 corresponds to iron and nickel, the most stable elements, and this stability is also reflected in the abundant presence of these elements in the inner core of the earth. From Figure 4.13, it is seen that the maximum occurs at about mass number 56 (iron). Thus, the nucleus of iron is thermodynamically most stable. It is also seen that the points for helium (mass number = 4), carbon (mass number = 12) and oxygen (mass number = 16) lie quite high in the graph. This shows that the nuclei of these elements are exceptionally stable. The maximum binding energy is seen to be about 8.7 MeV. This is the energy required to remove a proton or a neutron from the most stable nucleus.

4.2.7 Nuclear Reactions Nuclear reactions involve changes in the number of nucleons present in the nucleus. Hence, there is a formation of new atomic species. Thus, nuclear reactions lead to atomic transformations. In addition, a huge amount of energy is released in nuclear reaction, involving a small but measurable loss in mass. This mass is transformed into energy in accordance with Einstein’s equation (E = mc2). Fe

Binding energy per nucleon (MeV)

9 8

16O

4He

12C

6 5 4 3 2 1 0

75 100 125 150 175 200 225 250 Mass number

Figure 4.13 Mean binding energies of various nuclei

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4.16 Engineering Chemistry Rutherford carried out the first nuclear reaction by bombarding nitrogen (target nucleus) with a-particles (projectile) to produce an oxygen isotope. 14 7N

Nitrogen

4 2 He

→

Helium (a -particle)

17 8O

Oxygen

1 1H

Proton

Types of Nuclear Reactions (i) Nuclear fission: For example, 235 92 U

+ 01 n →

140 93 1 56 Ba + 36 Kr + 3 0 n +

Enormous amount of energy

(ii) Nuclear fusion: For example, 2 1H

+ 21 H →

14 2 He +

Enormous amount of energy

Nuclear Fission Very high nuclei have a lower binding energy per nucleon than the nuclei with intermediate mass. Thus, the former are less stable than the latter. During 1934–1938, Otto Hahn and F. Strassmann observed that when 235U is bombarded with slow-moving thermal neutrons (energy = 0.025 eV), it undergoes fission giving barium (z = 56) as one of the products of fission. In order to explain this observation, L. Meither and O.R. Frisch suggested that after the capture of a neutron, the uranium nucleus gets excited and then splits into two fragments of approximately equal mass, which is known as nuclear fission. There is an invariably mass defect during fission, that is, the total mass of products of fission is less than the total mass of neutron and the 235U atom. The loss of mass appears in the form of energy according to Einstein’s mass-energy relation, E = mc2. The fission reaction is represented as follows: 235 92 U

+ 01 n →

(

235 92 U

)→

140 93 1 56 Ba + 36 Kr + 3 0 n +

Huge amount of energy

Excited atom

The splitting of a heavier atom like Uranium-235 into a number of fragments of much smaller mass by suitable bombardment with sub-atomic particles with liberation of huge amount of energy is called nuclear fission. In this reaction, in addition to the release of enormous amount of energy, the most significant feature of nuclear fission is that more neutrons are produced than those consumed in the reaction. This process is schematically represented in Figure 4.14. Mechanism of Nuclear Fission—the Liquid Drop Model In the fission process, a heavy nucleus splits apart into nearly two equal fragments of more stable nuclei of intermediate mass. This fission reaction that occurs due to absorption of neutron may be treated as analogous to the behaviour of a liquid drop as it contracts, elongates and eventually splits apart into two

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Alternate Energy Resources

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Figure 4.14 A schematic diagram representing the uranium fission process

Figure 4.15 Schematic diagram representing the fission process by the Bohr-Wheeler liquid drop model for heavy nuclei droplets as shown in Figure 4.15. According to this liquid-drop model, the absorption of neutron by a nucleus causes the nucleus to oscillate like a liquid drop and then split’s it into two small, stable nuclei. There are 14 isotopes of uranium, and their mass number ranges from A = 227 to A = 240. The most important isotopes are 235U and 238U whose natural relative abundance is 0.72% and 99.28%, respectively. Both 235U and 238U undergo fission upon absorbing a neutron. 238U undergoes fission with ‘fast’ neutrons while 235U with both ‘fast’ and ‘slow’ neutrons. Among naturally occurring nuclides, only 235U undergoes fission, but 238U and 232Th are converted into 239Pu and 233U by neutron capture and two successive b-decay. These two converted nuclides then undergo fission. Both 235U and 233U can be excited to fissionable state by slow neutrons more easily than 238U as both are less stable. The division of 235U occurs in different ways and nearly 34 elements were identified in fission products. Enormous amount of energy is also liberated. In any single reaction, two particular nuclides are produced along with two or three secondary neutrons. These neutrons have about 200 MeV of kinetic energy. The daughter nuclei produced in fission reactions have different Z and A values. Some of many pathways through which 235U undergoes fission with fast neutrons are mentioned here.

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4.18 Engineering Chemistry

235 92

U+ 0n

236 92

160 62

Sm +

146 57

72 30

Zn + 4 0 n + energy 1

La +

87 35

Br + 3 0 n + energy

140 56

Ba +

93 36

Kr + 3 0 n + energy

144 55

Cs +

90 37

Rb + 2 0 n + energy

Xe +

90 38

144 54

1 0

Sr + 2 n + energy

Since emission of more than one neutron occurs in the fission process, there is a possibility of a chain reaction in which the releases of energy increases in geometric fashion. The neutrons released in one fission reaction cause a second fission reaction and so on, and a chain reaction continues, which, if goes unchecked, would quickly lead to the release of enormous amount of energy. A chain reaction is possible only when the amount of fissionable material exceeds its critical mass (Figure 4.16). The critical mass is defined as the amount of fissionable material, which is just large enough to recapture one neutron, on an average, for every fission reaction. If the amount of fissionable material is less than the critical mass, less than one neutron is recaptured; the rate of fission events does not grow, and the rate of energy release is low. In addition, if the amount of fissionable matter is more than the critical mass, the number of fission events increases, and a chain reaction is set up, which quickly grows into explosive proportions. Release of Fission Energy On this basis, nuclear fission reactions may be categorised into two types: (i) Uncontrolled nuclear fission used for construction of nuclear weapons, the atomic bombs (ii) Controlled nuclear fission, which is exploited for the controlled generation of energy in nuclear reactors 133 57 Sm 147 54 Ba

1 n 0

235 U 92

1 n 0

235 U 92

150 Xe 57

1 n 0 1 n 0

1 n 0 101 Nb 43

235 U 92

1 n 0

235 U 92

1 n 0 90 Sr 36

235 U 92

Figure 4.16 Self-propagating nuclear chain reactions

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Alternate Energy Resources

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(i) Uncontrolled nuclear fission—the atom bomb: In a fission reaction, two or three neutrons are produced. These neutrons can collide with other fissionable atoms to sustain and multiply the fission reactions. If the amount of fissionable material exceeds the critical mass, an uncontrolled explosive chain reaction may result. An enormous amount of energy is produced in the chain reaction. A nuclear bomb is a frightening example of the enormous amount of energy released by nuclear fission, which is not controlled. The central feature of a fission explosion is a growing chain of fission reactions. For this, there are three requirements as follows:

(a) The fissionable nuclide must be concentrated enough so that it becomes critical. (b) The sub-critical portions of this fissionable nuclide must be combined into a critical mass. (c) The critical mass must be held together for a long period so that the chain multiplies to immense size. The potential of uncontrolled nuclear fission was first realised in the atomic bomb (Figure 4.17). It contains two sub-critical portions of fissionable materials. One portion is propelled into another to form supercritical mass by carefully designed detonation of an ordinary chemical explosive such as trinitrotoluene (TNT). The fission chain multiplies, and then a nuclear fission explosion occurs. Tremendous amount of heat energy and many other radio nuclides are also released; their effects are disastrous to life and environment. The radioactive dust and debris are called fall-out. In 1945, the United States started the nuclear age by dropping two nuclear bombs on Hiroshima and Nagasaki, Japan. Both these bombs were fission weapons of tremendous power. The bomb, which was dropped on Hiroshima, contained 235U (Uranium-235) while the Nagasaki bomb had 239 Pu (Plutonium-239) as fissionable materials.

Subcritical235U

Subcritical235U TNT explosive charge

Figure 4.17 A simple design of atomic bomb (ii) Controlled nuclear fission—nuclear reactor: For controlled release of fission energy, the chain reaction is carried out in a device called a nuclear reactor. The fission is controlled in such a manner that on an average, only one neutron is left from each fission; to excite further fission, the large amount of energy released in nuclear fission can be used to generate electrical power. This requires a delicate balance between neutron generation and neutron loss, and this is achieved by proper use of moderator and control rods in the nuclear reactors.

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4.20 Engineering Chemistry Two types of nuclear reactors are in common use. They are as follows: (a) Thermal reactors: Most thermal reactors in the United States are ‘light water reactors’ in which ordinary water is used as moderator to slow the neutrons. However, heavy water reactors had been developed in Canada in which heavy water, D2O is used as moderator in place of ordinary water. Light water reactor (LWR), which is a type of thermal reactor, uses normal water as its coolant as well as neutron moderator. There are three varieties of LWRs: (1) Pressurised water reactor (PWR) (2) Boiling water reactor (BWR) (3) Supercritical water reactor (SCWR) • Reactor design: The LWR produces heat by controlled nuclear fission. The nuclear reactor core is the major portion of a nuclear reactor where the nuclear reactions take place (Figure 4.18). It consists essentially of the following parts: • Fuel: The fissionable material used in the reactor is called fuel. The fuel used is enriched uranium-235 (in the form of U3O8). This is obtained from the naturally occurring U-235. The solid fuel is made into the form of rods or pellets, shielding by placing them in stainless steel tubes. • Moderator: The most efficient fission reactions occur with slow neutrons. Thus, moderators, which are atoms of comparable mass, slow down the fast neutrons produced in fission and do not absorb them. Light water reactor uses ordinary water, also called light water, as its neutron moderator (Figure 4.18). • Control rods: To control the fission process, rods made of cadmium or boron are suspended between the fuel rods. These rods can be raised or lowered and control the fission process by absorbing neutrons. Therefore, they are called control rods. Cadmium and boron are good neutron absorbers. 113 48 Cd 10 5B

+ 01 n → 114 48 Cd +g

+ 01 n → 115 B+g

Containment structure Steam line Pressure vessel

Turbine generator

Isolation valves

Steam

Pump

Core

Control rods Water pool

Condenser cooling water

Figure 4.18 Light water nuclear reactor

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Alternate Energy Resources

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• Cooling system: The LWR uses ordinary water to keep the reactor cooled. The light water is circulated in the reactor core to absorb the heat, which is generated due to fission reaction. This transfers the heat to a steam generator, which converts water into steam, and this steam is then taken to turbines, which drive generators to produce electricity. • Shielding: To prevent loss of heat and protect the persons operating the reactor from radiation and heat, the entire reactor core is enclosed in a steel containment vessel. This vessel is housed in a thick-walled concrete building. The operating people are protected by a thick layer of organic material made of compressed wood fibres, which absorb the neutrons, b particles and g-rays. (b) Breeder reactors: Considering the rate at which U-235 is being used to produce power, the stocks are likely to exhaust very soon. Therefore, scientists have been actively engaged in investigating other fissionable materials. They have found plutonium-239 (Pu-239) and uranium-233 (U-233) to be quite suitable. These are produced by bombardment of more abundantly available U-238 and Th-232 with neutrons. A breeder reactor is a nuclear reactor that breeds fuel. It consumes fissile and fertile material at the same time as it creates new fissile material. In this reactor, neutrons produced from fission of U-235 are partly used up for carrying on fission of U-235 and partly to produce Pu-239 or U-233. Such reactors produce more fissionable materials (as Pu-239 or U-233) than they consume (as U-235). The sequence of reactions that take place when U-238 and Th-232 are bombarded with fast neutron producing the fissionable nuclei of Pu-239 and U-233 are represented as follows: (1) Fissionable Pu-239 by neutron bombardment followed by two successive b-decays. 238 92 U

+ 01 n →

239 92 U

−

b 239 b→ 239 93 Np → 94 Pu

(2) Fissionable U-233 by neutron bombardment on Th-232 by two successive b-decays. 232 90Th

+ 01 n →

233 90Th

−

b 233 b→ 233 91 Pa → 92 U

The neutrons produced by fission reactions are absorbed in a blanket of uranium or thorium. Since higher temperature is required to operate, water cannot be used as a coolant. Liquid sodium is used as a coolant. This type of reactor uses fast neutrons; therefore, no moderator is required. The nuclides such as U-238 and Th-232, which can be converted into fissionable nuclides, are called fertile nuclides, whereas nuclides such as U-235 and Pu-239, which are fissionable and are called fissile nuclides. Nuclear Fusion Nuclear fusion may be defined as a process of combination of two lighter nuclides to form a heavier nuclide with the release of energy. The following reactions have been successfully investigated as fusion reactions:

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1 2 1H + 1H

→ 23 H + γ

2 2 1H + 1H

→ 42 H + Q(= 3.25 MeV)

2 2 1H + 1H

→ 31 H + 11 H + Q(= 4.0 MeV)

2 3 1H + 2 H

→ 42 He + 11 H + Q(=18.3MeV)

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4.22 Engineering Chemistry These processes are generally known as thermonuclear reactions, because they require that the colliding nuclei must possess very high kinetic energies before they are initiated. The high kinetic energies of the reacting nuclei overcome the coulombic repulsion between positive particles of reactants. This is only possible of extremely high temperature of millions of degrees (≈ 4 × 106 °C), so these processes generally occur in the sun and other stars because of tremendous high temperatures that are nearly impossible to achieve and contain on earth. It is, therefore, believed that in the sun, the following process takes place: 4 11 H → 42 He +2 +01 e + γ There is evolution of 26 MeV of energy, and this energy is available to us from the sun and keeps the sun at extremely high temperature. The fission reactions take place only with a few rare and extremely heavy nuclides. On the other hand, fusion reactions are possible for light nuclides such as 1H, which are abundant. Furthermore, fusion reactions release more energy per unit mass than fission reactions. For example, fission of U-235 yields 7.9 ×107 KJ/g energy while fusion of two isotopes such as hydrogen, deuterium and tritium releases 3.4 ×108 KJ/g energy. Another advantage of fusion reaction is that the fusion reaction produces radio nuclides of very short half lives, so there would be no long-term waste disposal problem. There are several ways to study fusion reactions as follows: (i) By using particle accelerators. (ii) Through stellar nuclear reactions, the process occurs in the sun and other stars. (iii) By fusion bombs, which use a fission bomb to produce a temperature high enough for fusion. Fusion Bombs (Uncontrolled Nuclear Fusion Reactions) A hydrogen bomb uses nuclear fusion. A conventional explosive first triggers a fission bomb, which then induces the fusion reaction. These bombs are more powerful than fission bombs because they can incorporate large masses of nuclear fuel to produce unlimited energy. A hydrogen bomb has an arrangement of nuclear fission in the centre, which is surrounded by a mixture of deuterium ( 21 H) and lithium 6 isotope ( 63 Li). Nuclear fission provides heat and neutrons. The neutrons are used up for converting lithium isotope into tritium ( 31 H), and the heat liberated is required for the fusion between 21 H and 31 H to start. The fusion reactions are then accompanied by the liberation of a large amount of energy. Thus the reactions taking place in a hydrogen bomb may be represented as follows: Fission (in the centre) → Heat Neutrons 6 3 Li

+ 01 n → 31 H + 42 He + 4.78 MeV

2 3 1H + 1H

→ 42 He + 01 n + 17.6 MeV

2 2 1H + 1H

→ 23 He + 01 n + 3.3MeV

3 3 1H + 1H

→ 42 He + 2 01 n + 11MeV

All these processes occur in an extremely short time to release an immense amount of energy, and the bomb blasts.

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Alternate Energy Resources

4.23

Controlled Nuclear Fusion The major problem in a nuclear fusion reaction is the attainment of high temperature required for the purpose. So far, attempts to maintain such a high temperature to generate energy on a large scale have been successful for only a very small fraction of a second, because all known containers would vapourise at such a high temperature. At this temperature, the atoms become fully ionised, and the ions (i.e., nuclei and electrons) form the plasma state in which the particles move about independently but in the form of inter penetrating gases. The plasma is electrically neutral. Lastly, it can be mentioned that the attempts to make fusion reactors have not met with any success so far because of many technical difficulties involved, for example, that of the container material, which can withstand very high temperatures as required for the fusion to start. However, if a solution is found, it will place at our disposal a tremendous source of energy at a very cheap rate because deuterium is present in huge amount in sea water (in the form of D2O). The differences between nuclear fission and nuclear fusion are listed in Table 4.1. Table 4.1 Differences between nuclear fission and nuclear fusion S.No.

Nuclear fission

Nuclear fusion

It involves breaking up of a heavier nucleus into lighter nuclei. It is a chain process. It is initiated by neutrons of suitable energy and does not need high temperature. It can be controlled, and the energy release can be harnessed for useful purpose. Large number of radio-isotopes is formed, and there is nuclear waste. It requires minimum size of fissionable material, and if the size of the material exceeds the critical size, the reaction becomes explosive.

It involves union of two lighter nuclei to form heavier nuclei. It is not a chain process. It is initiated by very high temperature, by a few million degrees. It is difficult to control this process.

2. 3. 4. 5. 6.

There is no nuclear waste in this process. There is no limit to the size of the fuel for the reaction to start. However, the fuel does not undergo fusion until heated to a very high temperature of a few million degrees.

4.3 REVIEW QUESTIONS 4.3.1 Fill in the Blanks 1. In wind, is converted into mechanical energy or electrical energy. [Ans.: kinetic energy] 2.

is used to produce biofuel. [Ans.: algae]

3. Geo thermal resources are . [Ans.: hydrothermal, geopressure, hot dry rocks and magma] 4. Complete the following reactions: (a) 7 N14 + 01 n → 147 C+ (b)

13 Al

→ 15 P30 + 01 n

[Ans. : a = 1 H1 ; b = 42 He]

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4.24 Engineering Chemistry 5.

is used as a coolant in nuclear reactor. [Ans.: light water]

6. Atom bomb is based on [Ans.: nuclear fission] 7.

isotope of uranium is used in nuclear reactor. 235

[Ans.: U ] 8. Nuclear reaction is produced large quantities of energy according to [Ans.: Einstein energy mass E = mc2]

equation.

9. A fission reactor which produces more fissionable material than its consumed in its operation is reactor. [Ans.: breeder] 10.

changes alter the number of protons and neutrons in the nuclei of atoms consumed while simple chemical changes involve in reorganization of only. [Ans.: nuclear, electrons]

11. Sun’s energy is given by fusion of [Ans.: hydrogen]

nuclei.

12. Photovoltaic cell converts directly incident solar radiations to [Ans.: electric current] 13. The major material used in pv cell is [Ans.: silicon]

4.3.2 Multiple-choice Questions 1. Generating mechanical power by using wind is called (a) Solar energy (b) Tidal energy (c) Wind energy (d) Biofuels [Ans.: d] 2. Geothermal energy comes from (a) Heat from within earth (c) Heat from organic matter [Ans.: a]

(b) Heat from atmosphere (d) None of the above

3. Bio-fuel producing pollutes are (a) H2 and O2 (c) H2O [Ans.: b]

(b) Oxides of sulphur and nitrogen (d) None of the above

4. Energy which is not harm to the environment is (a) Solar energy (b) Wind energy (c) Alternative energy (d) None [Ans.: c]

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Alternate Energy Resources

4.25

5. Among the following, which one belongs to alternative energy? (a) Tidal Energy (b) Wind energy (c) Solar energy (d) All of the above [Ans.: c] 6. The source of energy in nuclear fuel is due to (a) Chemical reaction (b) Nuclear reaction (c) Nuclear fission (d) Nuclear fusion [Ans.: c] 7. Binding energy of a nucleus is related with (a) ∆Mc2 (b) Mc2 (c) ∆Mc (d) hv [Ans.: a] 8. Fissionable material used in nuclear reactor is (a) U235 (b) U238 (c) Th232 (d) Pu239 [Ans.: a] 9. 21 H + 21 H → ? (a) 23 He (b) 42 H (c) 42 He (d) None of these [Ans.: c] 10. Control rods used in nuclear reactor is made of (a) Na (b) B (c) CO2 (d) U235 [Ans.: b] 11. Atom bomb is based on principle of (a) Nuclear fusion (c) Chemical reaction [Ans.: b]

(b) Nuclear fission (d) None of these

12. Uncontrolled nuclear fusion reaction takes place in (a) Atom bomb (b) H-bomb (c) Nuclear bomb (d) Radium bomb [Ans.: b] 13. Solar energy originates from the reaction taking place in the sun (a) Nuclear fusion (b) Nuclear fission (c) Chemical reaction (d) Nuclear reaction [Ans.: a] 14. Photovoltaic cells are commonly known as (a) Primary cell (b) Solar cell (c) Secondary cell (d) Fuel cell [Ans.: b]

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4.26 Engineering Chemistry 15. Example of indirect solar energy is (a) Nuclear energy (c) Chemical energy

(b) Wind energy (d) Surface energy

[Ans.: b]

4.3.3 Short Answer Questions 1. Give definitions of alternative energy from different sources. Ans.: Oxford definition: Energy fueled into ways that do not use up natural resources or harm the environment. Princeton WordNet: Energy derived from sources that do not use up natural resources or harm the nature. Natural resource defense council: Energy that is not popularly used and is usually environmental sound such as solar or wind energy. Material and management: Fuel source other than those derived from fossil fuels typically interchangeably used for renewable energy. For example, wind, solar, biomass, wave and tidal energy. Torridge District Council: Energy generated from alternatives to fossil fuel; but need not be renewable. Climate Change 2007: Energy derived from nontraditional sources. For example, compressed natural gas, solar, hydroelectric and wind energy 2. Write a short note on wind energy. Ans.: The term ‘wind energy’ or ‘wind power’ describes the process by which the kinetic energy of wind is used to generate mechanical power or electricity with turbines. 3. Explain geothermal energy. Ans.: Geothermal energy comes from the heat within the Earth. Geothermal is a Greek word; geo–earth and thermal–heat. Geothermal energy is used to produce electricity, green houses and to heat building for other sources. 4. Write a short note on biofuels. Ans.: Biofuel is any fuel derived frombiomass of recently living organisms or their metabolic byproducts, such as manure from cows. Algae fuel is a biofuel which is derived from algae. During photosynthesis, algae and other photosynthetic organisms capture carbon dioxide and sunlight and convert them into oxygen and biomass. This is usually done by placing the algae between two panes of glass. The algae create three forms of energy fuel: they are from its growth cycle heat can generate, the oil derived from the algae used as biofuel, and on maturity algae can give biomass.

4.3.4 Descriptive Questions Q.1 Write a brief note on geothermal energy. Q.2 Write a note on wind energy. Q.3 Explain the importance of biofuels.

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Alternate Energy Resources

4.27

Q.4 Explain any three alternative energy resources. Q.5 Define alternative energy resources and explain water power. Q.6 Discuss the theoretical principles involved in the generation of power by nuclear fission and nuclear fusion. Q.7 Describe the various components of a nuclear power reactor and their functions. Q.8 Discuss the environmental aspects of nuclear power generation. Q.9 Write informative notes on the following: (a) Breeder reactors (b) Energy from nuclear fusion. Q.10 Write notes on nuclear fission and fusion. Q.11 (a) Explain how fission grade U235 is obtained. (b) Write short note on nuclear binding energy. Q.12 What are the functions of following in a nuclear reactor? (a)

235

(b) Cadmium rods Q.13 Define nuclear reaction. Q.14 Calculate the binding energy in Mev of 42 H , if its experimentally determined mass is 4.00390 amu. The masses of a proton, an electron and a neutron are respectively 1.007825, 0.0005852 and 1.008668 amu. Q.15 Explain how binding energy is useful for the stability of a nucleus. Q.16 Write the difference between nuclear fusion and nuclear fission. Q.17 Write the principle involved in atomic bomb and its reactions. Q.18 Write the reaction taking place in sun and stars.

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Electrochemistry and Batteries

5.1 Introduction Electrochemistry is defined as the branch of chemistry which deals with the relationship between electrical energy and chemical changes taking place in redox reactions i.e., how chemical energy produced in a redox reaction can be converted into electrical energy or how electrical energy can be used to bring about a redox reaction which is otherwise non-spontaneous. When electricity is passed through the solution of an electrolyte to bring about a redox reaction known as electrolysis and the arrangement is called an electrolytic cell and when electricity is produced by redox reactions (which are spontaneous) gives rise to what are known as electrochemical cell or galvanic cell or voltaic cell. In an electrolytic cell, the flow of electricity through the solution is due to the flow of ions and in an electrochemical cell, the flow of current is due to flow of electrons in the external circuit and the flow of ions through the solution in the inner circuit. The flow of current due to movement of ions through the solution of an electrolyte is known as electrolytic conduction. Thus, the three main aspects of study in the branch of electrochemistry are: (i) Electrolysis (ii) Electrolytic conduction (iii) Electrochemical cells

5.2 Electrolysis Electrolysis is a process of decomposition of an electrolyte by the passage of electricity from an external source through its aqueous solution or molten state for performing chemical reactions. Electrolysis requires the use of electrolytic conductors (electrolyte) in the form of an aqueous solution or in the molten state as well as electronic conductors (electrodes) which are essential components of an electrolytic cell. Two metal electrodes are dipped in electrolyte and are connected to a source of electricity i.e. battery. The electrode, which is connected to a positive pole of the battery is called anode and second electrode is called cathode which is connected to negative pole of the battery. Oxidation and reduction occur at the electrodes as shown in Figure 5.1.

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5.2 Engineering Chemistry D.C. Source

Ammeter

e−

Anode

Cathode − − − − − −

+ + + + + +

Figure 5.1 Electrolytic cell When an electrolyte is dissolved in water or taken in the molten state is dissociate into charged ions i.e., cation and anion. On passing electric current, cations move towards the cathode and anions move towards anode. After reaching at their respective electrodes, reaction takes place. Oxidation occurs at the anode while reduction takes place at the cathode. (i) For example electrolysis of molten sodium chloride, it consist of Na+ and Cl− ions. NaCl(l ) Na + (l ) + Cl − (l ) On passing electricity, ions moves towards their respective electrodes. Sodium metal is liberated at the cathode and Cl2 is evolved at the anode. At cathode: 2 Na + + 2e − → 2 Na (reduction) At anode: 2Cl − → Cl 2 + 2e − (oxidation) The electrons released at the anode pass through the external circuit and reach the cathode so that Na+ ions can be reduced. Thus, on oxidation reaction occurs at the anode and a reduction reaction at the cathode. (ii) Electrolysis of an aqueous solution of copper sulphate in electrolytic cell using Pt electrodes, reduction of the cupric ions occurs to copper and is deposited on the cathode, simultaneously release of oxygen gas at the anode occurs and reaction may be represented as:

At cathode: Cu 2 + + 2e − → Cu (reduction) 1 − − At anode: 2OH → H 2 O + O2 + 2e (oxidation) 2 Both SO42- and OH− ions are present near anode. Since the discharge potential of OH− ions is lower than that of SO42- ions, therefore, OH− ions are discharged with respect to SO42- ions and SO42- anions are charge carriers in the electrolyte.

5.2.1 Laws of Electrolysis Faraday in 1833 put forward the relationship between the amount of a substance deposited or dissolved during the electrolysis of aqueous solutions and the quantity of electricity passed through the electrolyte in the form of two laws of electrolysis.

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Electrochemistry and Batteries

5.3

Faraday’s First Law of Electrolysis The first law of electrolysis states that the mass of any substance deposited or liberated at any electrode due to passage of an electric current is proportional to the quantity of electricity passed. If W gram of the substance is deposited on passing Q coulombs of electricity, then W ∝ Q or W = ZQ Where Z = constant of proportionality and is called as electrochemical equivalent (ECE) of the substance deposited. If a current C amperes is passed for t seconds, then Q=C×t W=Z×Q=Z×C×t

So that

Thus if Q = 1 coulomb or C = 1 ampere or t = 1 second, W = Z. Hence, electrochemical equivalent (ECE) of a substance may be defined as the mass of the substance deposited when a current of one ampere is passed for one second i.e., a quantity of electricity equal to one coulomb is passed. Electrochemical equivalent can be calculated from the equivalent weight, as one Faraday (96500 coulombs) deposits one gram equivalent of the substance, Z=

Eq. wt. of the substance 96500

Faraday’s Second Law of Electrolysis Faraday’s second law states that when the same quantity of electricity is passed through solutions of different electrolytes connected in series, the amounts of different substances produced or deposited at the electrodes are directly proportional to their equivalent weights. Thus if w1 and w2 are the amounts produced or deposited as the two substances having chemical equivalent weights of E1 and E2 on passing Q coulombs of electricity. w1 E1 Z1 = = w2 E2 Z 2 The charge carried by one mole of electrons can be obtained by multiplying the charge present on one electron with Avogadro’s number i.e. it is equal to (1.6023 × 10-19 coulombs) × (6.022 × 1023) = 96490 coulombs (∼96, 500 coulombs or 26.8 A.hr). This quantity of electricity is called one Faraday or Faraday’s constant and is represented by F. Hence, Faraday’s constant F = 96490 C mol −1 96500 C mol −1 (i) In terms of electrons: If n electrons are involved in the electrode reaction, the passage of n Faradays (i.e., n × 96500 C) of electricity will liberate one mole of the substance. (ii) In terms of gram equivalents: One Faraday’s (i.e., 96500 Coulombs) of electricity deposits one gram equivalent of the substance.

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5.4 Engineering Chemistry Thus by knowing the weight of substance deposited (w gram) on passing a definite quantity of electricity (Q coulombs), the equivalent weight of the substance can be calculated as: w × 96500 Q Q = c×t

Eq. wt. =

w × 96500. c×t So, by knowing the quantity of electricity passed, the amount of substance deposited can be calculated. So,

5.3 Electrolytic conDUCTION (i) The substances which allow electricity to pass through them are known as conductors. For example metals, graphite, acids, bases, fused salt etc. (ii) Some substances which do not allow electricity to pass through are known as insulators. For example: Mica, non-metals, wood, rubber, benzene etc. (iii) Conductors are further divided into two classes such that one class is for those which conduct electricity without undergoing any decomposition. These are called electronic conductors e.g., metals, graphite and other minerals. In such type of substances, conduction is due to the flow of electrons. (iv) The second class is for those which conduct electricity when current is passed through them or they undergo decomposition. These are called as electrolytic conductors or electrolytes e.g., solution of acids, bases and salt in water, fused salt etc. In such type of substances, conduction is due to the movement of ions. (v) There are two types of electrolytes such as strong electrolyte and weak electrolyte. Strong electrolytes are those which dissociate almost completely in the aqueous solution or in the molten state and conduct electricity to a large extent e.gs. → NaOH, KOH, HCl, HNO3, H2SO4 etc. and weak electrolytes are those which have low degree of dissociation and hence conduct electricity to a small extent e.gs. → NH4OH, CH3COOH, HCN, Ca(OH)2, Al(OH)3 etc. (vi) The substances which don’t dissociate and donot conduct electricity are known as non-electrolytes e.gs: Sugar, urea, glucose etc.

5.3.1 Factors Affecting Electrolytic Conduction (i) Viscosity of the solvent decreases with increase of temperature. Hence electrolytic conduction increases with increase of temperature. (ii) Polarity of the solvent affect on electrolytic conduction, greater is the polarity of the solvent, greater is the ionization and hence greater is the conduction. (iii) In case of concentrated solution, conduction is less but as the dilution starts, conduction is also increases. (iv) When temperature increases, dissociation of solution increases and hence the conduction also increases.

5.3.2 Electrical Resistance and Conductance Every substance offers resistance to the flow of electricity to a small or large extent. This law is known as Ohm’s law. This law is applicable to both metallic conductors as well as electrolytic conductors. Ohm’s law states that if to the ends of a conductor is applied a voltage ‘E’ and a current ‘I’ flows through it, then resistance ‘R’ of the conductor is given by

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Electrochemistry and Batteries

5.5

E I When current is measured in amperes, voltage is measured in volts. If one ampere current flows through a conductor when a voltage of one volt is applied to it, then resistance of the conductor is taken as 1 Ohm (written as 1 ‘Ω’ omega) E Volts R= or Ohm = I Amperes According to Ohm’s law, when a substance offers greater resistance will allow less electricity to flow through it. 1 I∝ R The reciprocal of the electrical resistance is called as conductance. It is usually represented as ‘C’. Thus 1 C= R Units 1 C= Ohm = Ohm −1 or mho or siemens i.e., (1 S = 1 Ω−1) The S.I unit of conductance is Siemens (S). R=

5.3.3 Specific, Equivalent and Molar Conductivities Specific Conductivity Specific conductivity is also known as conductivity. Experimentally, it is observed that resistance R of a conductor is (i) Directly proportional to its length (l) and (ii) Inversely proportional to its area of cross section (a) i.e., R∝

l a

R= r

l a

Where r is a constant of proportionality, called specific resistance or resistivity. Value of resistivity depends upon the material of the conductor. “The reciprocal of resistivity is known as specific conductivity or conductivity or electrolyte conductivity.” It is denoted by k (kappa) or K. Thus, if K is the conductivity and C is the conductance. Then, we know that 1 1 R= and r= C K ∴ Formula change into l a 1 1 l = C Ka R= r

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5.6 Engineering Chemistry Now if l = 1 cm and a = 1 cm 2, then K = C. Hence conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross section. Alternatively, it may be defined as conductance of one centimeter cube of the solution of the electrolyte as represented by the Figure 5.2. 1cm

−

1cm

Figure 5.2 Electrolyte conductivity “If the volume of the solution is V cm3, then conductivity of such a solution at this dilution V is written as KV.” Units Resitivity (r ) = R

a l

cm 2 cm = ohm.cm or Ω cm = ohm

Conductivity (K ) =

1 r

1 ohm.cm = ohm −1 cm −1 or Ω −1 cm −1 or mho cm −1 =

C.G.S unit of conductivity = mho cm−1 or S cm−1 S.I unit of conductivity = S m−1 1 S m−1 = 0.01 S cm−1

5.3.4 Equivalent Conductivity Consider the solutions having equal volumes and containing their corresponding gram equivalent weights for the comparison of the conductances of the solutions of different electrolytes. Then conductance of such solution is called its equivalent conductivity or equivalent conductance. Hence equivalent conductance of a solution is defined as the conductance of all the ions produced from one gram equivalent of the electrolyte dissolved in V cm3 of the solution when the distance between the electrodes is one cm and the area of the electrodes is so large that whole of the solution is contained between them. Equivalent conductivity or equivalent conductance is represented by Λeq (lambda) (i) If the volume of solution is V cm3 containing one gram equivalent of the electrolyte, Then, Equivalent conductivity = Specific conductivity × V Λeq = KV × V

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Electrochemistry and Batteries

5.7

(ii) If the solution has concentration of C gram equivalent per litre i.e., C gram equivalents are present in 1000 cm3 of the solution, then volume of solution containing one gram equivalent will 1000 be . C Hence, Λeq = K v × V can be written as 1000 1000 = Kc × Λ eq = K c × Ceq Noramality Units Λ eq = K × V = ohm −1 cm −1 ×

cm3 gram eq

= ohm −1 cm 2 (g eq) −1 or mho cm 2 eq −1 or S cm 2 eq-1 S.I unit is S m 2 eq-1 Hence,

1 S m 2 eq-1 = 104 S cm2 eq-1

5.3.5 Molar Conductivity The molar conductivity of a solution is defined in a manner similar to that of equivalent conductivity. The term molar conductivity is sometimes also called as molar conductance or molecular conductivity. It is usually represented as Λm or m. The molar conductivity of a solution is defined as the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of the electrodes is so large that whole of the solution is contained between them. Similar to equivalent conductivity, molar conductivity is given by following expression as follows: or

Λm = Kv × V Λ m = Kc ×

1000 1000 = Kc × Molarity C

Where Kc = is specific conductivity V = Volume of the solution containing one mole of the electrolyte C = molar concentration Units Λ m = Kv × V cm3 mol cm 2 mol −1 or Ω −1 cm 2 mol −1

= ohm −1 cm −1 × = ohm −1

or mho cm 2 mol −1 or S cm 2 mol −1

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5.8 Engineering Chemistry

5.3.6 Measurement of Electrolytic Conductance As we know that conductance is the reciprocal of the resistance. Hence conductance can be obtained by the measurement of the resistance and resistance can be found by the principle of Wheatstone bridge method. In finding the resistance of the solution of an electrolyte, a special type of cell has to be used such that the solution is present between the two electrodes. The cell thus used is called as conductivity cell. It consists of platinised Pt electrodes at a fixed distance apart and resistance is measured using alternating current to prevent any deposition on the electrodes during electrolysis as represented in Figure 5.3. In this process, an alternating current of frequency 500-2000 Hz is used. A signal generator such as a variable frequency oscillator, a null detector-indicator such as ear phone or a sensitive micro ammeter is used as a detector. Variable resistance R

Cell

Jockey

B Electrolytic solution

Earphone

Alternation current source

Conductivity Cell

Figure 5.3 Wheatstone bridge circuit Hence, Wheatstone bridge circuit consists of (i) (ii) (iii) (iv)

Source of alternating current which is either from induction coil or a vacuum tube oscillator. An earphone Platinised Pt electrodes Variable resistance with standard resistance box

In this method, a suitable value of resistance R is introduced from the standard resistance box such that when the sliding contact i.e., the Jockey J is moved along the stretched wire, the sound in the earphone is reduced to minimum at the point somewhere in the middle of the wire AB, say at the point C. Then if X is the resistance of the electrolytic solution, then by Wheatstone bridge principle, Resistance R Resistance of wire AC = Resistance X Resistance of wire CB length AC = length BC length BC Resistance, X = Resistance R × length AC Thus, knowing the resistance R and the balance point C, resistance X of the electrolytic solution can be calculated.

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Electrochemistry and Batteries

5.9

Hence, Conductance, C =

1 1 length AC = × Resistance X R length BC

To calculate specific conductivity, which is related with the conductance according to the expression K =C ×

l a

Thus specific conductivity of a solution can be determined by measuring its conductance and the distance (l) between the electrodes and area (a) of cross section of each of the electrodes. l For a particular cell, is constant and this constant is called as cell constant. a Hence, Specific conductivity (K) = conductance × cell constant Thus, the cell constant of any particular cell can be found by measuring the conductance of a solution whose specific conductivity is known. First of all we have to remove all types of organic as well as other oily impurities which are sticking to the walls of the cell and electrodes by cleaning with dilute N chromic acid. Then we dipped the electrodes of the cell in KCl solution whose specific conductivity 50 is known. N By Kohlrausch it is verified that at 25 °C value of specific conductivity of KCl is 50 0.002765ohm-1cm−1. Hence by knowing the value of cell constant, the specific conductivity of the given solution can be determined by measuring its conductance and multiplying the value with the cell constant. Equivalent conductivity (Λ eq ) can be calculated by using the relation Λ eq = Specific conductivity × =K×

1000 Ceq

1000 1000 =K× Ceq Normality

Molar conductivity (Λm) can be calculated by using the relation 1000 C 1000 1000 =K× =K× C Molarity

Λ m = Specific conductivity ×

Hence, by knowing the molar concentration, C and specific conductivity K, Λeq (equivalent conductivity) and Λm (molar conductivity) can be calculated.

5.3.7 Variation of Conductivity with Concentration As we know that, specific conductivity (K) as well as equivalent conductivity (Λeq ) and molar conductivity (Λm) vary with the concentration of the electrolyte. The specific conductivity K decreasing with increasing dilution (decrease in concentration) of the electrolyte solution while equivalent and molar conductance increases with increase in dilution (decrease in concentration) which is easily understood by the relation.

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5.10 Engineering Chemistry Equivalent conductivity Λ eq = K ×

1000 = K ×V C

and Molar conductivity Λ m = K ×

1000 = K ×V C

As mentioned above, equivalent conductance increases with increase in dilution and this value approaches to a limiting value as the dilution of the electrolytic solution increases. This limiting equivalent conductance value is called the equivalent conductance at zero concentration or infinite dilution Λ0 or Λa . With increasing dilution, dissociation of the electrolyte increases, hence equivalent conductance increases but the number of ions per unit volume decreases so specific conductance K decreases. On the basis of an empirical relationship between the equivalent conductance and concentration given as Λc = Λ0 − b c Where b is a constant depending upon the nature of the solvent. Λ0 = specific conductivity at infinite dilution This equation is called Debyl Huckel-Onsagar equation and is found to hold good at low concentrations.

5.3.8 Conductance Behaviour of Strong Electrolyte In case of strong electrolyte, the equivalent conductance does not vary very much with dilution. A linear graph is obtained for low concentrations but it is not linear for higher concentrations. The curve shows that there is small increase in conductance with dilution. This is because a strong electrolyte is completely dissociated in solution and so the number of ions remains constant with increase in dilution, conductance increases and approaches a maximum limiting value at infinite dilution i.e., Λ0 or Λ∝. Examples of strong electrolytes include strong acids and bases (NaOH, HCl, H2SO4, and KOH) and solutions of ionic solids (NaCl, KNO3, KCl etc).

5.3.9 Conductance Behaviour of Weak Electrolyte In case of weak electrolyte, equivalent conductance increases rapidly with decrease in concentration. But conductance of a weak electrolyte is much lower than that of a strong electrolyte at the same concentration. Further the curve obtained for a weak electrolyte shows that there is a very large increase in conductance with dilution especially near infinite dilution as shown in Figure 5.4. This is because as the concentration of the weak electrolyte is reduced, more of it ionizes. Thus, increase in conductance with decrease in concentration is due to the increase in the number of ions in the solution. However, it does not reach a limiting value, So Λ0 or Λ∝ cannot be determined graphically as in the case of strong electrolyte. Examples of weak electrolytes include weak acids and bases (acetic acid, other organic acids and ammonia) and weakly dissociating salts.

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Electrochemistry and Batteries

Equivalent conductance Λeq.(mho cm2 mol−1)

Λ0

5.11

Strong electrolyte (KCI)

Weak electrolyte (CH3COOH)

Concentration( c ) (mol L−1)−½

Figure 5.4 Variation of equivalent conductance of a strong and weak electrolyte with electrolyte concentration Variation of equivalent conductance with concentration of strong (KCl) and weak (CH3COOH) e lectrolytes at 25 °C Equivalent conductance mho cm2 mol-1 Concentration C mol L-1 0.10 0.05 0.02 0.01 0.001 0.0005

Strong electrolyte KCl

Weak electrolyte CH3COOH

128.6 133.4 138.3 141.3 146.9 149.86

5.2 7.4 11.6 16.2 48.6 390.71

5.4 Kohlrausch’s Law Of Independent Migration Of Ions The equivalent conductance of an electrolyte solution is equal to the sum of the conductivity of the constituent cations (l +) and anions (l-) and is expressed as Λeq = l + + lThe equivalent conductance of an electrolyte solution increases with increasing dilution. At high concentrations, the greater inter-ionic attraction retard the motion of ions and therefore the conductance falls with increasing concentrations. But with decrease in concentration (increase in dilution) the ions are far apart and therefore the interionic attractions decreases due to which the conductance increases with dilution. So, at infinite dilution, the equivalent conductance reaches its maximum value as the retarding effects of the ionic atmosphere completely disappear. Λ 0 = l 0c + l 0a

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5.12 Engineering Chemistry Where l 0c and l 0a are called the ionic conductivities at infinite dilution for the cation and anion respectively and this equation is known as the Kohlrausch’s law of independent conductance of ions. The equivalent conductivity of an electrolyte at infinite dilution is the sum of two values one depending upon the cation and the other upon the anion. This equation stated as at infinite dilution, each ion makes a definite contribution to the equivalent conductance of the electrolyte whatever be the nature of the other ion of the electrolyte. In 1875, Kohlrausch made a series of measurement in which he observed that the difference between Λ° values for each pair of sodium and potassium salts having a common anion was same, irrespective of what this anion was. Similarly, the difference in the Λ° values for each pair of salts having the different anions and a common cation was same, irrespective of what this cation was. Λ° values at 25 °C of some pair of electrolytes having common ions Electrolyte KCl

Λ0

Difference

149.86

Electrolyte

Λ0

KBr

151.92

KCl NaBr

149.86 128.51

NaCl LiBr

126.45 117.09

LiCl

115.03

23.41 NaCl KBr

126.45 151.92

NaBr KNO3

128.51 144.96

NaNO3

121.55

Difference 2.06

23.41

2.06

23.41

2.06

The ionic conductivity values of some common ions at 25 °C is presented in Table 5.1. Table 5.1 Ionic conductivities at infinite dution at 25 °C is in mho cm2 Cations +

H Li+ Na+ K+ Ag+ NH +4 Mg2+ Ca2+ Ba2+ Cu2+ Zn2+ Sr2+

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l0c

Anions -

l0a

349.8 38.6 50.1 73.5 61.9 73.3

OH FClBrINO3-

198.6 55.4 76.4 78.1 76.8 71.5

53.1

ClO4-

67.4

59.5 63.6 53.6 52.8 59.4

IO HCOOCH3COOSO42CO32-

54.5 54.6 40.9 80.0 69.3

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Electrochemistry and Batteries

5.13

Applications of Kohlrausch’s Law Conductance measurements have a wide range of numerous applications. A few of them are discussed below: (i) Determination of equivalent conductivity at infinite dilution (L0) for weak electrolytes: As already mentioned, the equivalent conductivity of a weak electrolyte at infinite dilution cannot be determined experimentally, firstly because the conductance of such a solution is low and secondly because the dissociation of such an electrolyte is not complete even at very high dilutions. The equivalent conductivity at infinite dilution can be calculated using Kohlrausch’s law. Consider the example of acetic acid (CH3COOH) as the weak electrolyte. By Kohlrausch’s law Λ0 (CH3COOH) = l0 (CH3COO-) + l0(H+) This equation can be arrived at by knowing the molar conductivities at infinite dilution for the strong electrolytes KCl, CH3COOK and HCl as by Kohlrausch’s law Λ0 (KCl) = l0 (K+) + l0 (Cl-) Λ0 (CH3COOK) = l0 (CH3COO-) + l0 (K+) l0 (HCl) = l0(H+) + l0(Cl-) Hence, we required l0 (CH3COO-) + l0(H+) = l0 (CH3COO-) + l0(K+) + l0(H+) + l0(Cl-) - l0(K+) - l0 (Cl-) i.e. Λ0 (CH3COOH) = Λ0 (CH3COOK) + Λ0(HCl) - Λ0(KCl) (ii) Determination of the degree of dissociation or ionisation of weak electrolyte: According to Arrhenius theory of electrolytic dissociation, the increase in the equivalent conductivity with dilution due to the increase in the dissociation of the electrolyte and reaches upto a limiting value at infinite dilution. Thus if ΛC is the equivalent conductivity of a solution at any concentration and Λ0 is the equivalent conductivity at infinite dilution (i.e., zero concentration). So, Degree of dissociation (a ) =

ΛC Λ0

The value of Λ0 for the weak electrolyte can be calculated by using Kohlrausch’s law. (iii) Determination of dissociation constant of a weak electrolyte: Knowing the degree of dissociation (a), the dissociation constant (K) of the weak electrolyte at concentration (C) can be calculated using the formula: − + CH 3 COOH CH 3 COO + H

Initial concn n

Equilibrium conc When C moles are taken

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1-a C(1-a)

a Ca

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5.14 Engineering Chemistry Dissociation constant (K ) = =

[CH 3 COO − ][H + ] [CH 3 COOH ] Ca ⋅ Ca Ca 2 = C (1 − a ) 1 − a

Thus by knowing degree of dissociation (a), dissociation constant (K) can be easily calculated. (iv) Determination of ionic product of water: It is found that ionic conductances of H+ and OH- at infinite dilution are l0(H+) = 349.8 ohm-1 cm2 l0(OH-) = 198.6 ohm-1 cm2 By Kohlrausch’s law Λ0(H2O) = l0(H+) + l0(OH-) = 349.8 + 198.6 = 548.4 ohm-1 cm2 Specific conductance of pure water at 25 °C found to be, K = 5.54 × 10-8 ohm-1 cm-1 Applying the formula, 1000 C K × 1000 5.54 × 10 −8 × 1000 = C= Λ0 548.4

Λ0 = K ×

= 1.01 × 10 −7 g ion mool −1 ∴ Kw = [H+] [OH-]

= (1.01 × 10-7) × (1.01 × 10-7)

= 1.02 × 10-14 Here, Kw is ionic product of water. (v) Determination of solubility of sparingly soluble salts: Salts such as AgCl, BaSO4, PbSO4 etc which dissolve to a very small extent in water are called sparingly soluble salts. As they dissolve very little, their solutions are considered as infinitely dilute. Further as their solutions are saturated, their concentration is equal to their solubility. Thus, by knowing the specific conductance (K) and equivalent conductance (Λ) of such a solution, we have Λ0 = K ×

1000 1000 =K× solubility C

Λ0 = K ×

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1000 S

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Hence, Solubility = K ×

Electrochemistry and Batteries

5.15

1000 Λ0

Since the solutions are extremely dilute, the conductance contribution of water is also considered. So we have to be subtracted the value of water from the total conductance. Therefore S=

( K − K ω ) × 1000 Λ0

Where Kw is the specific conductance of pure water. And value of Kw is 1.60 × 10-6 S cm-1 at 25 °C. Ionic Mobility The ionic mobility is defined as the velocity with which an ion moves under a potential gradient of 1volt per cm in a solution. Ionic mobility =

Velocity of the ion Field strength

Hence, Units of ionic mobility: In CGS system: cm 2 s-1 v-1 In S.I system: m 2 s-1 v-1 Ionic mobility at infinite dilution (u0) is related to ion conductance at infinite dilution l0 and it is obtained by dividing the equivalent conductance of the ion by the Faraday. l i.e., Ionic mobility l0 = u0 F ⇒ u0 = 0 F Hence, Ionic mobility of cation (u+ ) =

l+0 F

Ionic mobility of anion (u− ) =

l−0 F

and

Where l+0 and l−0 are the equivalent conductance of the cation and anion of the electrolyte.

5.5 Conductometric titrations Conductometric titration is a simple and accurate technique used in volumetric analysis to determine the end point of a titration. Principle Conductometric titration is based on the fact that the conductance of a solution at a constant temperature depends upon the following points:

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5.16 Engineering Chemistry (i) The number of ions present in solution (ii) Charge on the ions in solution (iii) Mobility of the ions present in solution Conductance change during titration because in titration process one type of ions are replaced with other kind of ions which differ in their mobilities. At the end point, there is a sharp change in conductance. Process In this process, titrant is added from a burette into a solution taken in a beaker. Conductivity cell is dipped into this beaker by which conductance is measured after every addition of a titrant. Conductance of a solution is plotted against volume of titrant added by taking conductance along y-axis and volume of titrant along x-axis. Two straight lines are plotted and end point is the point of intersection of two straight lines. Burette Titrant solution Conductivity cell

Solution

Figure 5.5 Arrangement for conductometric titration Types of conductometric titrations: (i) Acid-base or neutralization titrations (ii) Displacement titrations (iii) Precipitation titrations and complexation titrations (i) Acid-base titrations: (a) Titration of a strong acid with a strong base: Reaction between a strong acid (HCl) with a strong base (NaOH) is written as + − + − + − H + Cl Na + OH Na + Cl + → + H2O

Strong acid

Strong base

Strong salt

By taking acid solution (HCl) in a beaker and base (NaOH) into the burette, conductance of 20 ml of HCl is measured with conductometer on every 1 ml addition of NaOH solution from the burette. A graph is plotted by taking volume of base along x-axis and conductance along y-axis.

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5.17

From the graph, it is clear that conductance first decrease due to replacement of fast moving H+ ions with slow moving Na+ ions till the end point. After end point, conductance sharply increases due to presence of fast moving OH- ion in the solution. A +

]d ec H

sin

Conductance

a re

−]

g sin

rea

inc

B O

End point

Volume of NaOH added

Figure 5.6 Titration curve of strong acid vs strong base (b) Titration of a strong acid with a weak base: The reaction between a strong acid with a weak base is written as + − H + Cl → NH +4 + Cl − + H 2 O + NH 4 OH 

Strong acid

Weak base

Strong salt

When a strong acid is titrated against a weak base, a graph as shown below is obtained. From the graph, it is clear that conductance first decrease sharply due to replacement of fast moving H+ ions with slow moving NH4+ ions. After the end point, there is very small increase in conductance because of presence of weak base in the solution.

Conductance

O End point

Volume of NH4OH added

Figure 5.7 Titration curve for strong acid vs weak base

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5.18 Engineering Chemistry (c) Titration of a weak acid with a strong base: The reaction between a weak acid with a strong base is written as + − CH 3 COOH + Na + OH → CH 3 COO − + Na + + H 2 O 

Weak acid

Strong base

Strong salt

Conductance

During titration of a weak acid with a strong base, a graph as shown below is obtained. From the graph, it is clear that conductance first decrease due to formation of strong salt (CH3COONa) having a common ion effect (CH3COO –) which suppresses the ionization of weak acid. There is a small increase in conductance up to end point, due to formation of strong salt (CN3COO – Na+) which completely ionizes in the solution. After the end point, conductance sharply increases due to presence of strong base (Na+OH–) in the solution.

O End point

Volume of NaOH added

Figure 5.8 Titration curve of weak acid vs strong base (d) Titration of a weak acid with a weak base: The reaction between a weak acid (CH3COOH) with a weak base (NH4OH) is written as CH 3 COOH + NH 4 OH  → CH 3 COO − + NH +4 + H 2 O Weak acid

Weak base

Strong salt

From the graph, it is clear that conductance first decreases due to formation of common ion (CH3COO –), which suppresses the dissociation of weak acid. Further increase in conductance up to end point is observed due to formation of strong salt (CH3COONH4) which completely dissociates into ions. After the end point, there is small increase in conductance due to presence of weak base (NH4OH) in the solution.

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Conductance

5.19

O End point

Volume of NH4OH added

Figure 5.9 Titration curve of weak acid vs weak base

lv H aO sN

OOH

CH 3C x

vs Na

Conductance

(e) Titration of a mixture of strong acid and weak acid with a strong base: In such titrations of a mixture of strong acid (HCl) and weak acid (CH3COOH) with a strong base (NaOH), the graph obtained is shown hereunder. When NaOH solution is added to the mixture of acids, the strong acid (HCl) is neutralized first due to high ionization. Two end points ‘x’ and ‘y’ are obtained and point ‘x’ corresponds to neutralization of strong acid (HCl) with a strong base (NaOH). Point ‘y’ corresponds to neutralization of weak acid (CH3COOH) with a strong base (NaOH). After end point ‘y’, there is sharp increase in conductance due to presence of strong base (Na+OH–) in the solution.

Volume of NaOH added

Figure 5.10 Titration curve of strong and weak acid vs strong base (ii) Displacement titrations: Upon addition of HCl into sodium acetate (CH3COONa), displacement reaction taken place and is written as

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5.20 Engineering Chemistry + − + − CH 3 COO − Na + + H Cl → CH 3 COOH + Na + Cl 

Strong salt

Strong acid

Strong salt

Conductance

From the graph, it is clear that there is small increase in conductance up to end point, which is due to replacement of highly dissociated CH3COONa with strongly dissociated NaCl and undissociated CH3COOH. After the end point, conductance increases due to presence of HCl in the solution.

O End point

Volume of HCl added

Figure 5.11 Displacement reaction of CH3COOH vs HCl (iii) Precipitation titrations: Consider a precipitation reaction like AgNO3 vs KCl. The precipitation reaction is written as Ag + + NO3− + K + + Cl −  → AgCl ↓ + K + + NO3− ppt .

Conductance

From the figure, it is clear that conductance remains constant up to end point due to same mobilities of Ag+ and K+ ions. After the end point, there is a sharp increase in conductance due to presence of free ions (K+ and Cl-) in the solution.

O End point

Volume of KCl added

Figure 5.12 Precipitation reaction of AgNO3 vs KCl

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5.21

Advantages of Conductometric Titrations (i) Conductometric titrations are more accurate. (ii) They can be used for titrating colored solution because the colored solutions cannot be titrated by ordinary volumetric methods using indicators. (iii) They can be employed even for titrating very dilute solutions. (iv) They can be used for titrating weak acids and weak bases. (v) No observation is required near the end point because end point is detected graphically. Limitations of Conductometric Titrations (i) It can be applied only to a limited number of titrations. (ii) When the total electrolytic concentration is high, the results from conductometric titration become less accurate, less precision and less satisfactory.

5.6 Electrochemical cells A device used to convert the chemical energy produced in a redox reaction into electrical energy is called an electrochemical cell or simply a chemical cell. Electrochemical cells are also called galvanic cells or voltaic cell after the names of Luigi Galvani and Alessendro Volta, who were the first to perform experiments on the conversion of chemical energy into electrical energy. The galvanic cell consists of two half-cells combined in such a way that oxidation takes place in one beaker and reduction takes place in another beaker and both two electrodes are connected externally by a piece of metal wire and an electric current flows through the external circuit. The practical application of galvanic cell is Daniel cell as shown in Figure 5.13. It consists of a zinc rod dipped in zinc sulphate solution taken in a beaker and a copper rod is placed in copper sulphate solution taken in another beaker. The two portions of the cells are called half cells or redox couples in which oxidation half reaction in one beaker and reduction half reaction in another beaker. The two electrodes are connected by a wire and two solutions are connected by a salt bridge. Voltmeter / Ammeter e−

Flow of electrons

2−

SO4

Anode (−)

e− Salt bridge(K2SO4)

K+

Zinc electrode

Movement of cations Zn2+

Cathode(+) Copper electrode

2−

SO4 Movement of anions

ZnSO4 soln.

CuSO4 soln.

Figure 5.13 Daniel cell

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5.22 Engineering Chemistry Salt bridge is an inverted U-shaped tube containing concentrated solution of an inert electrolyte like KCl, KNO3, K 2SO4 etc., or solidified solution of such an electrolyte in agar-agar and gelatine. The inert electrolyte does not take part in the redox reaction. The main functions of the salt bridge are (i) To complete the electrical circuit by allowing the ions to flow from one solution to the other without mixing of the two solutions. (ii) To maintain the electrical neutrality of the solutions in the two half-cells. The oxidation and reduction reactions that occur at the two electrodes may be represented as: At the anode: Zn → Zn 2+ + 2e- (Oxidation half reaction) At the cathode: Cu2+ + 2e- → Cu (Reduction half reaction) The overall cell reaction is: Zn + Cu2+ → Zn2+ + Cu or Zn + CuSO4 → ZnSO4 + Cu Some important features of the electrochemical cell may be summed up as follows: (i) The zinc electrode at which oxidation takes place is called the anode. The copper electrode at which the reduction takes place is called the cathode. (ii) Due to oxidation of zinc electrode, electrons are produced at the zinc electrode and it pushes the electrons into the external circuit and hence it is designated as negative pole. The other electrode required electrons for the reduction of Cu2+ into Cu. Therefore, it acts as the positive pole. (iii) The electrons flow from the negative pole to the positive pole in the external circuit and conventional current is flowing in opposite direction. (iv) The oxidation of zinc into ions produces excess of Zn2+ ions in the left beaker. Similarly, reduction of copper ions to copper leaves the excess of SO42- ions in the solution in the right beaker. (v) To maintain electrical neutrality of the solution in the two beakers, the cations and anoins move through the salt bridge which helps to complete the inner circuit. Representation of a Galvanic Cell An electrochemical cell is represented in a manner as illustrated below for the Daniel cell: Zn ZnSO 4 (C1 ) CuSO 4 (C2 ) Cu or Zn Zn 2+ (C1 ) Cu 2+ (C2 ) Cu (i) The electrode on which oxidation takes place is written on the left hand side and other electrode on which reduction takes place is written on right hand side. (ii) Anode is written by writting the metal first and then the electrolyte and cathode is written by first writing the electrolyte and then metal. (iii) Single vertical lines represent the phase boundaries of the electrodes and double line represents the salt bridge.

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5.23

Electrode Potential Consider a metal rod (M) placed in contact with a 1 M solution of its own ions (M n+) at 25 °C, then there are possibilities such as (i) The metal atoms of the metal rod (M) may lose electrons and changed into Mn+ ions, i.e., metal atoms get oxidized M → Mn+ + ne- (Oxidation) (1) (ii) The Mn+ ions, on collision with the metal rod may gain electrons and changed into metal atoms i.e., Mn+ ions are reduced. Mn+ + ne- → M (Reduction)

(2)

What actually happens depends upon the relative tendency of the metal or its ions. If metal has relatively higher tendency to get oxidized then reaction (1) will occur. If the metal ions have relatively higher tendency to reduced, then reaction (2) will occur. During oxidation negative charge is developed on metal rod and during reduction positive charge develops on metal rod. Metal rod (M)

M+ne−

Metal rod (M)

M M

M+ne−

Thus in either case, there is a separation of charges between the metal rod and its ions in the solution. As a result, a potential difference exists between them. “The electrical potential difference set up between the metal and its ions in the solution is called electrode potential or the electrode potential may be simply defined as the tendency of an electrode to lose or gain electrons when it is in contact with solution of its own ions”. The electrode potential is termed as oxidation potential if oxidation takes place at the electrode with respect to standard hydrogen electrode and is called as reduction potential if r eduction takes place at the electrode with respect to standard hydrogen electrode. The electrode potential is called standard electrode potential if metal rod is suspended in a solution of one molar concentration and the temperature is kept at 298 K. Measurement of Electrode Potential The absolute value of the electrode potential of a single electrode potential cannot be determined because oxidation half reaction or reduction half reaction cannot take place alone. It can only be measured by using some electrode as the reference electrode. The reference electrode used is the standard or normal hydrogen electrode (S.H.E or N.H.E). It consist of platinum foil or wire coated with platinum black dipped into a molar (1M) solution of H+ ion and hydrogen gas at 1 atmospheric pressure is continuously passed through it at 298 K. This electrode may serve as anode or cathode depending upon the nature of another electrode to which it is connected.

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5.24 Engineering Chemistry Pt Wire

H2 gas (1 atm. pressure) Glass tube

Hg Platinised pt foil

1M HCl soln. Standard hydrogen electron

The reaction, when electrode acts as the anode i.e., oxidation takes place H2(g) → 2H+(aq) + 2eWhen this electrode act as the cathode i.e., reduction takes place 2H+(aq) + 2e- → H2(g) This electrode is usually represented as Pt, H2(g), (1 atm)/H+(1M) The electrode potential of the standard hydrogen electrode is taken as 0.000 at 298 K. The standard electrode potential of the other electrode can be determined by connecting it with S.H.E and finding EMF of the cell experimentally. As the EMF of the cell is the sum of oxidation potential where oxidation takes place and reduction potential of the electrode where reduction takes place and as the electrode potential of S.H.E is zero, so EMF of the cell will directly gives the electrode potential of the electrode under investigation. EMF or Cell Potential of a Cell As we know that electrochemical cell is made up of two electrode i.e., two half cells. One of these electrodes must have a higher electrode potential than the other. As a result of this, the electrons flow from an electrode at a higher potential to the electrode at a lower potential. The difference between the reduction potentials of the two half cells is known as electromotive force (EMF) of the cell or cell potential or cell voltage.

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EMF = Reduction potential of cathode – Reduction potential of anode EMF = Ecathode – Eanode Ecell = Eright – Eleft

The EMF of the cell depends on (i) The nature of the reactions (ii) Concentration of the solution in the two half cells reactions (iii) Temperature. EMF Measurement EMF of any electrochemical cell is determined by potentiometric method. The measurement becomes possible when the positive pole of the cell is connected to the end B of the potentiometer wire and the negative pole to the sliding contact. If the connections are wrong the balance point cannot be determined and hence the polarity of the electrodes is also indicated by the circuit. The emf of the cell (Ex) is measured by comparing with the emf of a standard cell (Es), Whose EMF is accurately known and remains constant at a given temperature. The sliding contact jockey J is moved along the wire AB till there is no deflection (zero current flow) in the galvanometer (point C on the wire). The emf of the standard cell (Es) is proportional to the length AB and emf of the cell (Ex) is proportional to the length AC and the emf of the unknown cell Ex is calculated by the formula for no deflection in the galvanometer G. Es length AB = Ex length AC or length AC Ex = Es × length AB

−

Standard cell (Es) +

Jockey

B Stretched wire

−

Experimental cell (Ex)

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5.26 Engineering Chemistry Electrochemical Series As we know that emf or cell potential or cell voltage can be calculated as Ecell = Ecathode - Eanode or = Eright - Eleft If the electrode potential or half - cell potential are measured at 25 °C (298K), then E°Cell = E°Cathode - E°anode E°Cell = standard cell potential In the Daniel cell, E°Cell = E°Cu

/Cu

- E°Zn

/Zn

Where E°Cu /Cu and E°Zn /Zn are the standard reduction potential for copper and zinc ion. (Always remember that when no specific mention is made the electrode potential is always the reduction potential.) “Electrochemical series is a series in which various electrodes have been arranged in order of their increasing values of standard reduction potential”. Electrochemical series as shown in Table 5.2. 2+

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Electrode reaction Li+ + e → Li K+ + e- → K Ca+2 + 2e-→ Ca Na+ + e- → Na Mg2+ + 2e- → Mg Al3+ + 3e-→ Al Zn2+ + 2e-→ Zn Fe2+ + 2e-→ Fe Cd2+ + 2e-→ Cd Sn2+ + 2e → Sn Pb2+ + 2e → Pb Fe3+ + 3e-→ Fe 2H+ 2e-→ H2 Sn4+ + 2e-→ Sn2+ Cu2+ + 2e-→ Cu Fe3+ + e-→ Fe2+ Ag+ + e-→ Ag + Cr2O27 + 14H + 6e → 2Cr3+ + 7H2O Ce4+ + e- → Ce3+ F2 + 2e- → 2F-

E° (Volts) -3.045 -2.925 -2.87 -2.714 -2.37 -1.66 -0.763 -0.44 -0.403 -0.136 -0.126 -0.036 0.000 +0.15 +0.337 +0.771 +0.799 +1.33 +1.61 +2.87

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Applications of Electrochemical Series (i) To compare the relative oxidizing and reducing powers: In an electrochemical series, the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative. The Li, Li+ electrode has the least E° Value and therefore, it is reduced with most difficulty. Therefore, Li is the strongest reducing agent and the species which are placed below hydrogen are easily reduced and their standard reduction potential values are positive. The F2, 2F- electrode has the highest E°value and therefore, F2 has the greatest tendency to get reduced, so F2 is the strongest oxidizing agent. (ii) To compare the relative activities of metals: Lesser the reduction potential of a metal, more easily it can lose electrons and hence greater is its reactivity. So, as a result, a metal with less reduction potential can displace metals with higher reduction potentials from their salt solutions. For example, Reduction potential of Mg, Zn, Fe, Cu and Ag are in the order: Mg < Zn < Fe < Cu < Ag. Hence, each metal can displace metals on its right from the salt solutions. (iii) To predict whether a metal reacts with acid to give hydrogen gas: Metal (M) may react with an acid to give hydrogen gas (H2), the following reaction takes place. 1 M + H+ → M + + H2 2 Which can split into two half reactions as: M → M+ + e−

(oxidation half reaction )

1 H ( reduction half reaction) 2 2 Thus, the metal should have the tendency to lose electrons i.e., undergo oxidation, w.r.t hydrogen so; the metal should have a negative reduction potential. Thus, all the metals above hydrogen in electrochemical series react with the acid to give hydrogen gas. (iv) To predict the spontaneity of any redox reaction: For a spontaneous reaction, EMF of the cell must be positive and EMF can be calculated by using the formula. Ecell = Ecathode - Eanode H+ + e− →

(v) To determine the equilibrium constant: As we know that ∆G° = -RT ln Keq and ∆G° = − nFE° RT ln Keq nF 2.303RT E° = log Keq nF 0.0591 E° = log Keq at 298 K n Therefore, by measuring of E° helps us to determine the equilibrium constant for the electrode reaction. E° =

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5.28 Engineering Chemistry Nernst Equation for Electrode Potential Nernst equation tells us the effect of electrolyte concentration and temperature on the electrode potential. For this purpose, the electrode reaction is written as reduction reaction. Mn+ (aq) + ne- → M(s) Since ∆G: −nFE and ∆G° = −nFE° Hence E = E° − 2.303

RT [M] log nF [M n+ ]

Where E = electrode potential under given concentration of M n+ ions and temperature T E° = Standard electrode potential R = gas constant T = Temperature in K F = 1 Faraday For pure solids or liquids or gases at one atmospheric pressure, the molar concentration is taken as unity [M] = 1 RT 1 E = E° − 2.303 log So, nF [M n + ] Putting R = 8.314 JK-1 mol-1 F = 96500 coulombs T = 298 K 0.0591 1 We get E = E° − log n [M n + ] In case of an electrochemical cell, aA + bB ne  → cC + dD −

Then applying Nernst equation, we have Ecell = E°cell −

[C]c [D]d 0.0591 log n [A ]a [B]b

Where n is the number of electrons involved in the cell reaction.

5.7 Types of electrodes Different types of single electrodes other than the one seen in Daniel cell are also presents in an electrochemical cell. Half cell reactions of such types of electrodes are discussed by using Nernst equation. Single electrodes may be grouped into the following types: (i) Metal-metal ion electrode: Such type of electrode consist of a pure metal (M) is dipped in a solution of its cation (M n+) The reaction may be represented as M M n+ + ne −

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Thus single electrode potential by Nernst equation is given by E = E° −

0.0591 [M n + ] log [M] n

Since [M] = 1 in pure solids or liquids form. Hence, E = E° − or

0.0591 log[ M n + ] n

E M/M n+ = E° M/M n+ −

0.0591 log[ M n + ] n

Ex.: When Zinc rod is dipped in ZnSO4 or copper rod is dipped in CuSO4 soln. (ii) Metal-amalgam electrode: Such type of electrode is set up when metal – amalgam (i.e., When metal dissolved in mercury to form metal – amalgam) is in contact with a solution of metal ion (M n+). Generally, more active metals such as sodium metal is used for metal – amalgam solution. The reaction may be represented as: M(Hg) M n+ + ne − and electrode potential by Nernst equation E = E° − or

0.0591 [M n + ] log [M( Hg)] n

E M(Hg) / M n+ = E° M/M n+ −

0.0591 [M n + ] log [M( Hg)] n

Where E°M/M is the standard potential of pure metal, [M(Hg)] is the activity of the metal in amalgam which is not unity. n+

(iii) Metal-metal insoluble salt electrode: Such type of electrode is set up when metal (M) is in contact with sparing soluble salt (MX) and dipped in a solution containing a common anion (X-). The reaction is represented as M(s) | MX | X- (aq) Electrode reaction as: M(s) + X − (aq ) MX(s) + e−

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5.30 Engineering Chemistry Electrode potential as by Nernst equation E = E° − 0.0591 log

1 [X − ]

or 1 [X − ] Ex.: (i) Silver – silver chloride electrode in which silver metal is dipped AgCl solution and then KCl solution Ag|AgCl|Cl-. (ii) Calomel electrode in which mercury is in contact with solid mercurous chloride and a solution of KCl. Hg|Hg2Cl2|ClE M/MX/X − = E° M/MX/X − − 0.0591 log

(iv) Gas electrode: Such type of electrode is set up when inert metal (e.g. Pt) dipped in a solution containing ions to which the gas is reversible and then gas is continuously bubbled through the solution. Example: Hydrogen electrode consist of a platinum wire coated with platinum black and dipped in a solution of H+ ions through which hydrogen gas is bubbled. The hydrogen electrode is represented as Pt|H2|H+ and electrode reaction is represented as 1 H (g) H + (aq ) + e − 2 2 and electrode potential is given by E = E° − 0.0591 log

[H + ] [H 2 ] 12

Since activity of [H2] =1

So E = E° - 0.0591 log[H+] + We know that pH = -log[H ] Hence E = E° + 0.0591 pH (v) Redox electrode: Such type of electrode is set up when inert metal (e.g., Pt) dipped in a solution containing common ions in two oxidation states of the substance. Example: W hen Pt wire is in contact with common ions in different oxidation state such as Fe2+ and Fe3+ Electrode is represented as Pt|Fe3+, Fe2+ Electrode reaction is represented as Fe3+ + e − Fe2 + Electrode potential is given by E = E° − 0.0591 log

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[Fe2 + ] [Fe3+ ]

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5.8 REFERENCE ELECTRODE As we know that, each electrochemical cell is made up of two electrodes. It is not possible to determine experimentally the potential of a single electrode. Electrodes whose potentials are exactly known and can be used for the construction of the electrochemical cell and by which we can determine the single electrode potentials are called as reference electrodes. The common examples of reference electrodes used include the standard hydrogen electrode (SHE), calomel electrode and silver-silver chloride electrode. References electrode are broadly classified into two types: (i) Primary reference electrode (ii) Secondary reference electrodes (i) Primary reference electrode: Standard hydrogen electrode (S.H.E) is used as a primary reference electrode because its standard potential is taken as zero at all temperature. But it is not always convenient to use standard hydrogen electrode because it is difficult to maintain the activity of H+ ions in the solution at unity and also to keep the pressure of the gas uniformly at one atmosphere. So, for these reason, some secondary reference electrodes like Ag − AgCl, calomel electrode, Quinhydrone electrode etc. are used. (ii) Secondary reference electrodes: They include Ag – AgCl electrode, glass electrode, calomel electrode, Quinhydrone electrode, their standard potentials are accurately determined and they are generally used in place of standard hydrogen electrode. (a) Standard calomel electrode (SCE): Calomel electrode is commonly used as a secondary reference electrode for potential measurements Calomel electrode consist of mercury, solid mercurous chloride and a solution of potassium chloride. The electrode is represented as Hg, Hg2Cl2(s); KCl solution. Construction of calomel electrode: It consist of pure mercury (Hg) placed at the bottom of a glass tube having a side tube on each side. Mercury (Hg) is covered with a paste of mercurous chloride Hg2Cl2 (Calomel), as shown in diagram. After that a solution of potassium chloride (KCl) is placed over the paste through the right side tube and the solution is also filled along the left side also after that a platinum wire is dipped into glass tube to make electrical contact of the electrode with the circuit as represented in Figure 5.14. Working of calomel electrode: (1) If reduction occurs on calomel electrode then reactions may be represented as follows: Hg 22 + (aq ) + 2e − 2Hg(l ) Hg 2 Cl 2 (s) Hg 22 + (aq ) + 2Cl − (aq ) It results into increase in the concentration of chloride ions in solution. (2) If oxidation occurs on calomel electrode then reactions may be represented as follows: 2Hg(l ) Hg 22 + (aq ) + 2e − Hg 22 + (aq) + 2Cl − (aq ) Hg 2 Cl 2 (s)

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5.32 Engineering Chemistry

Pt wire

Saturated KCl solution

Hg2Cl2 paste

Figure 5.14 Saturated calomel electrode It result into decrease in the concentration of Cl- ions and increase the Hg 2+ 2 ions in the solution. Thus, in case of the calomel electrode, the electrode reaction is generally represented as 1 Hg Cl (s) + e − Hg(l) + Cl − (aq ) 2 2 2 Electrode potential is given by E = E °Cl-/Hg

2Cl 2/pt

– 0.0591 log [Cl-]

Thus, potential of calomel electrode depends upon the concentration of chloride Cl- ions. The reduction potential of calomel electrode also varies with the concentration of KCl solution and reduction potential of the calomel electrode at 298 K for various KCl concentrations are on hydrogen scale are discussed below in Table 5.3: Table 5.3 Electrode potential of KCl solution at different concentration at 298 K [KCl]

Name

0. l N l.0 N Saturated

DCE Decinormal calomel electrode NCE Normal calomel electrode SCE saturated calomel electrode

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Electrode potential (Volt) 0.3335 0.2810 0.2422

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To obtain the potential of any other electrode it is combined with the calomel electrode and the emf of the resulting cell is measured. By which we can easily measure the potential of other electrode. Advantages (i) Calomel electrode is simple to construct. (ii) It does not vary with temperature. (iii) It is stable for a long time.

(b) Quinhydrone Electrode: This electrode is also used as a reference electrode. This is a redox electrode in which oxidation reduction takes place simultaneously. It consists of a platinum wire dipped in a solution containing equimolar ratio of hydroquinone (QH2) and Quinone (Q). The electrode reaction is represented as: O + 2e− + 2H+

Quinone (Q) (C6H4O2)

OH Hydroquinone (QH2) (C6H6O2)

The electron is represented as Pt, Q, QH2; H+ (aq)

Hydroquinone is reversible with [H+] concn The electrode potential at 298 K is given by E = E°pt / Q , H+, QH − 1

0.0591 log[QH 2 ] 2 [Q][H + ]2

Since, concentration of Quinone and hydroquinone is unity because both are taken in equimolar ratio. Hence,

E = E°PtQ, H+, QH − 2

0.0591 1 log + 2 2 [H ]

0.0591 log[H + ]2 2 0.0591 = E° + 2 × log[H + ] 2 = E° + 0.0591 loog[H + ]

E = E° +

As we know that pH = −log[H+]

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5.34 Engineering Chemistry So, E − E° − 0.0591 pH Quinhudrome (QH2) is used for the measurement of the pH of the solution. This electrode is not suitable for alkaline medium The standard electrode potential of the quinhydrone electrode, E° = +0.06996 V Hence, E = 0.06996 – 0.0591 pH Thus potential of quinhydrone electrode, depends upon the pH of the solution.

5.9 ION SELECTIVE ELECTRODES (ISE) An ion selective electrode consists of specially prepared membranes placed between two electrolytes and having the ability to respond to certain specific ions. So, it is also called as specific ion electrode (SIE). In such type of electrode the potential developed across the membrane which is related to the activities of the specific ion dissolved in a solution and this potential is measured by potentiometric device like a voltmeter or pH meter.

5.9.1 Electrochemical Circuit and Working of ISE The ISE consist of a tube, in which one end of the tube is fused to an electrically conducting membrane and the tube contains a gel incorporating the ion to which the electrode is sensitive and inert electrolyte such as potassium chloride. A silver wire in contact with the gel together with the inert electrolyte constitutes the internal silver-silver chloride reference electrode. After that ion selective electrode is coupled to a SCE and immersed in sample solution. The potential difference developed across the membrane and this potential difference is related with the activity of ions present in gel as well as in sample solution as shown in Figure 5.15. Voltmeter

Standard calomel electrode (SCE)

Ag-AgCl electrode

Ion selective electrode

External reference electrode

Membrane Sample solution

Figure 5.15 Electrochemical circuit

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The cell is represented as Membrane Reference electrode 1

Sample solution concentration = c1

Internal standard solution concentration = c2

External reference electrode

Reference electrode 2

Ion-selective electrode

The potential difference developed across the membrane is given by EISE =

C 0.0591 log 2 C1 n

and emf of the cell is given by Ecell = EISE − ∆E ref . (i) When same reference electrodes are used then ∆Eref = 0 (ii) When different reference electrodes are used then ∆Eref = constant (k) Suppose in a cell the reference electrode is cathode and so, ISE is cathode. Ecell = k −

C 0.0591 log 2 C1 n

Here, k is a constant depends upon the internal and external reference electrode and C1 and C2 are the concentration of the external and internal solution respectively.

5.9.2 Types of Ion – Selective Membranes There are four main types of ion – selective membranes used in ion selective electrode (i) (ii) (iii) (iv)

Glass membranes Solid state membranes/crystalline membranes Gas sensing membranes Liquid ion-exchange membranes

(i) Glass membranes: The electrode which is having glass membrane is very highly selective for some cations such as Na+, NH4+, Ag+ and Li+ and also selective for some double-charged metal ions, such as Pb2+ and Cd 2+. Glass membranes are made from ion-exchange type of glass containing Na 2O, CaO and SiO2 as shown in Figure 5.16(a). (ii) Solid state membranes/Crystalline membranes: This type of membrane is selective for both cation and anion of the membrane forming substance. This type of membrane is made up from lanthanum trifluoride (LaF3) Crystal doped with europium difluoride (EuF2) is sensitive for Fluoride F- ion and selectivity for other halides, cyanide, silver, lead, membranes is formed by pressing pellet with pure silver sulphide(Ag2S).

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5.36 Engineering Chemistry For example, for chloride (Cl-) ion, pellet of (AgCl & Ag2S) is pressed together within membrane as shown in Figure 5.16(b). (iii) Gas sensing membranes: The electrode having gas sensing membrane is used to measure the concentrations of dissolved gases such as carbon dioxide (CO2), ammonia (NH3), Sulphur dioxide (SO2) Nitrogen oxide (NO2) and oxygen (O2). Gas molecules diffuse across the membrane until the gas concentration are the same in the internal electrolyte and the sample solution. Any change in the gas concentration in internal electrolyte brings about a change in pH of the electrolyte and this pH is measured by glass electrode. For example, The CO2 gas sensing electrode has a sodium hydrogen carbonate (NaHCO3) solution as the internal electrolyte and cell reaction is CO2 (sample) + 2H2 O H 3 O+ + HCO3− (internal electrolyte) The concentration of ( HCO3− ) bicarbonate ion is considered as constant in the internal electrolyte. The pH of the glass electrode is a function of dissolved carbon dioxide in the sample solution. (iv) Liquid ion-exchange membranes: Such type of membrane is usually consisting of a large organic molecule capable of specifically interacting with an anions or cations. For cations such as calcium dialkyl phosphoric acid in which calcium chloride is taken as an internal solution. For alkali and alkaline earth metal cation, membrane is made up from phosphate diesters and neutral monocyclic crown ethers. For anions such as NO3− , ClO 4− , BF4− membrane is made up from tris–1,10-Phenanthroline 2+ Fe (ClO-4)2

Reference electrode Internal electrolyte

Glass membrane (a)

Figure 5.16 (a) Ion selective electrode of glass membrane

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Reference electrode

Internal electrolyte

(b)

Solid membrane

(c)

Liquid membrane

Figure 5.16 (b) Ion selective electrode of solid membrane and (c) Ion selective electrode of liquid membrane

5.9.3 Applications of Ion Selective Electrodes Ion selective electrodes have become extensively important in recent years, because of the fact that the potential of these electrodes solves a large number of practical problems. They have been widely used in clinical, biological, water, air, oceanographic and pharmaceutical research, and in general analytical determinations. These are commercially available and reliable for H+, NH3, F−, Cl−, Br−, I−, Cd 2+, CN−, − BF4, Pb+2, NO3−, ClO 4− , K+, Ag+, S2−, Na+, SCN−, SO2 and a variety of enzymes. The electrodes have been used for the following individual measurements and titrations: (i) It is possible to determine lead poisoning in blood and urine samples by atomic absorption or ashing the sample and using a colourimetric reagent for the lead in the residue. Lead can be measured directly in blood or in urine samples with a PbS/Ag2S electrode. No pre-treatment or separation is required. (ii) Chloride ion can be determined in a variety of industrial and physiological samples by making use of chloride electrode. Rapid accurate clinical determination of Cl− ion in sweat is an example. (iii) The distillation and titration procedure in Kjeldahl method may be avoided by using ammonia electrode. Nitrogen is converted into NH +4 ion and the solution is made basic and the concentration of NH3 is determined with ammonia electrode. (iv) Calcium electrode has been used to determine Ca+2 ion in beer, boiler water, soil, milk, minerals, serum, sea water, sugar, wine, etc. Ca2+ is one of the most important electrolytes in human physiology. The determination of Ca+2 in biological fluids and related samples is, thus, very important. Successful measurements of this type have been made with calcium ion exchange electrode and flow through electrode. The latter electrode is ideally suited for serum and other biological fluids because of increased selectivity of Ca2+ over Na+ and K+. (v) The electrode has also been used for measuring stability constants of Ca2+ complexes and to follow the kinetics of complex formation. The potentials of each electrode can be expressed by the Nernst equation. For example, for a reaction OX + ne − Red The Nernst equation may be written as E = E 0 − [ RT nF ]ln[Red ] [OX ]

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5.38 Engineering Chemistry where E0 is the standard potential of the reaction recorded on the scale in which the normal hydrogen electrode is zero. [OX] and [Red] are molar concentrations of the oxidised and reduced species, respectively. Actually, activities should be used rather than concentration. However, for potentiometric titration close to the end point, activity charges are very close to the concentration charges, and hence, concentrations are used for simplicity.

5.10 GLASS ELECTRODE Principle: pH of an aqueous solution depends upon the concentration of H+ ions and pH is determined by using glass electrode. When glass electrode is immersed in solution, whose pH value is to be determined, the potential difference develops across the membrane and this potential difference is proportional to the concentration of H+ ions.

5.10.1 Construction of Glass Electrode The glass electrode consists of an electrically conducting glass membrane which is made up from Na2O, CaO and SiO2. Glass electrode taken in the form of bulb and then sealed to the bottom of a glass tube as shown in Figure 5.17. The bulb contains 0.1M HCl solution and internal reference electrode such as Ag-AgCl electrode or platinum wire is dipped in solution to make electrical contact with the solution. Electrode is represented as Ag |AgCl(S) |(0.1M) HCl| glass

Reference electrode Ag-AgCl(s)

HCl (0.1 M) Glass membrane

Figure 5.17 Glass electrode Theory The glass membrane of glass electrode undergoes on ion-exchange reaction with the Na+ ions of the membrane with H+ ions. M − − Na + + H + M − − H + + Na + Membrane (M)

Membrane (M)

The potential difference developed across the gel layer of glass membrane between the two liquid and this potential difference developed due to the concentration (C1) of acid solution inside the bulb and concentration (C2) of acid solution into which glass bulb is dipped.

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RT C2 ln nF C1

EG = As we known that C1 = 0.1M

EG = E°G − 0.0591 log[H + ] As we know that pH = -log[H+] EG = E°G + 0.0591 pH E°G is glass constant For the measurement of pH of any unknown solution, the glass electrode is immersed in that solution, and then it is combined with a reference saturated calomel electrode. The electrochemical cell is represented as Ag |AgCl|0.1 MHCl| Glass |Solution of unknown pH| saturated calomel electrode EMF of the cell at 298 K is given as Ecell = Eright - Eleft = Eref - EG = Eref - EG Ecell = E ref − E°G − 0.0591 pH pH =

E ref − E°G − E cell 0.0591

EMF of calomel electrode at standard condition is Eref = 0.2422 pH =

0.2422 − E°G − E cell 0.0591

The value of E°G is obtained by measuring with the solution of known pH and EMF of such a cell is determined by a potentiometer. So, we can easily determine the pH of unknown solution cell arrangement is represented in Figure 5.18 for unknown solution. Potentiometer Pt wire Glass electrode

Solution of unknown pH 0.1 M Ag-AgCl(s) HCl electrode

KCl saturated solution Hg Hg + Hg2Cl2

Calomel electrode

Figure 5.18 Glass electrode-calomel electrode cell arrangement for pH determination

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5.40 Engineering Chemistry Advantages of Glass Electrode (i) (ii) (iii) (iv) (v) (vi)

It can be used even in strong oxidising solutions as well in alkaline solutions. The results obtained are quite accurate. It is simple to operate and most convenient. It is not easily poisoned. pH is easily determined with few milliliters of solutions. It can easily used in the presence of metallic ions.

Limitations of Glass Electrode The glass electrode is sensitive zeta ions such as Na+ in addition to H+, particularly at pH > 9 which result into the alternation of the linear relationship between pH and emf of the glass electrode.

5.11 CONCENTRATION CELL In concentration cell, EMF arises due to transfer of matter from one half-cell to the other because of a difference in the concentration of the species involved in two half-cell. Concentration cells may be classified into two types (i) Electrode – concentration cells (ii) Electrolyte – concentrations cell Concentration cells

Electrode concentration cells

Electrolyte concentration cells

Without transference

With transference

(i) Electrode concentration cells: These cells consist of two like electrodes at different concentration are dipping in the same solution of the electrolyte. For example, two hydrogen electrodes at different gas pressures immersed in the same solution of hydrogen ions constitute an electrode – concentration cell. This cell is represented as Pt:H2(p1) | Soln of H+ ions (HCl) | H2(p2):Pt electrode reaction is represented as, H 2 ( p1 ) 2H + + 2e − (Oxidation ) 2H + + 2e − H 2 ( p2 ) Overall reaction

( Reduction )

H 2 (p1 ) H 2 (p2 )

According to nearest equation, emf of the cell is 25 °C is given by

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Ecell = E° −

5.41

p 0.0591 log 2 2 p1

E° for concentration cell is zero So, Ecell = 0.02955 log

p1 at 25°C p2

The process is spontaneous when expansion of hydrogen gas from pressure p1 at one electrode to pressure p2 at the other electrode it means p2 < p1. Another example of such type of cell is that of an amalgam with same metal at two different concentrations. Hg – Pb (C1), PbSO4 (solution), Hg – Pb (C2) The electrode reaction is written as: Pb(C1 ) Pb 2 + + 2e − (Oxidation )

L.H.E

Pb 2 + + 2e − Pb(C2 )

R.H.E Overall reaction

( Reduction )

Pb(C1 ) Pb(C2 )

EMF of the cell is given by E = E right − Eleft 0.0591 0.0591     log C2  −  E°Pb − log C1  =  E°Pb − 2 2     C1 C1 0.0591 log = = 0.02955 log C2 C2 2 If C1 > C2, then EMF is positive, it means the whole process is spontaneous. (ii) Electrolyte-concentration cells: In these types of cells two identical electrodes are dipped in two metal ions solution at different concentration. Such type of cells is represented as M, M n + (C1 )  M n + (C2 ).M Example: Zn, Zn 2 + (C1 )  Ζn 2+ (C2 ), Zn Here C1 and C2 are the concentration of metal ion (M n+) in the two electrolyte and these two electrolytes are separated from each other by salt bridge and C2 > C1 for a spontaneous process. R.H.E L.H.E Overall reaction

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M n + (C2 ) + ne − M(S) n+

M(S) M (C1 ) + ne n+

( Reduction ) −

(Oxidation )

M (C2 ) M (C1 )

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5.42 Engineering Chemistry EMF of the cell Ecell = E right − Eleft

Ecell

0.0591 1   0.0591 1  log  −  E° − log  =  E° − n C1  n C2    C 0.0591 log 2 = n C1

If (i) C1 = C2 EMF of the cell, E = 0, it means reaction is at equilibrium (ii) C2 > C1, EMF of the cell comes out to be positive, it means reaction is spontaneous in nature.

For example:

Zn  Zn 2 + (C1 )  Zn 2 + (C2 )  Zn

The cell reaction is Zn 2 + (C2 ) Zn 2 + (C1 ) EMF of the cell Ecell = 0.0591 log

C2 C1

5.12 POTENTIOMETRIC TITRATIONS Potentiometric titrations are those in which emf of any cell is determined by plotting a graph between variation of electrode potential versus volume of titrant added. They are generally are of three types: (i) Acid-Base titrations (ii) Oxidation-Reduction titrations (iii) Precipitation titrations (i) Acid-Base titrations: For determining the strength of acid solution, we want to titrate a solution of HCl against NaOH. Any type of electrode whose electrode potential is depends upon the concentration of H+ ions (e.gs, quinhydrone electrode, glass electrode) is dipped in the HCl solution and then that electrode is connected with a reference electrode (e.g calomel electrode, Ag–AgCl(S) electrode) to form a electro chemical cell. For example, suppose hydrogen electrode is used as H+ indicating electrode and a saturated calomel electrode is used as the reference electrode and then both electrodes are dipped in acid solution which is taken in beaker. The electrochemical cell is represented as pt , H 2 (1 atm), H + ( unknown concentration C)  KCl Solution; Hg 2 Cl 2 ( s), Hg EMF is measured by potentiometer which is connected to both the electrodes during the addition of alkali solution from a burette.

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Ecell = E right − Eleft = Ecalomel − E hydrogen

Ecell = 0.2422 − 0.0591 logg  H +  Ecell = 0.2422 + 0.0591 pH

(1)

E. Volts

By knowing the EMF of the cell, we can determine the pH of the acid solution. During the titration i.e., when we titrating acid solution (HCl) with titrant (alkali solution NaOH), concentration of H+ ion goes on decreasing i.e., pH of the solution goes on increasing. Hence according to equation-1 EMF of the cell goes on increasing. After that a graph is plotted against electrode potential versus volume of alkali (NaOH) added and we obtained a sigmoid curve in which end point is analysed. The titration curve is steep near the end point.

End point

Volume of NaOH

Figure 5.19 Potentiometric acid-base titration curve As it is clear from the Figure 5.19 that for accurate determination of end point, curve should be steep near the end point, more accurately we can determined the end point by plotting a curve ∆E between versus volume of NaOH added and end point is determined by drawing a vertical ∆V line from the peak to the volume axis as shown in Figure 5.20.

∆E ∆V

End point

Volume of NaOH

Figure 5.20 Determination of end-point in acid-base titration curve (ii) Oxidation-reduction titrations: The redox titrations are also carried out potentiometrically same as in acid-base titrations. In redox titrations electrode reversible with respect to H+ ions is replaced by an inert metal, such as Pt wire, immersed in a solution containing both the oxidized and reduced form of the same species.

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5.44 Engineering Chemistry The electrode acts as an oxidation-reduction electrode and this electrode is combined with a reference electrode, e.g., a saturated calomel electrode (SCE) to form a galvanic cell. The cell is represented as Hg, Hg 2 Cl 2 (s), KCl(sat.soln)  Fe2 + , Fe3+ ; pt The EMF of the cell is measured potentiometrically at each stage of titration and end point is obtained near the steep point of the curve as in case of acid-base titration curve. (iii) Precipitation titrations: In such type of titrations, potential of the half-cell is measured by connecting it with the calomel electrode. For example, precipitation reaction of silver nitrate (AgNO3) with potassium chloride (KCl) Ag + + NO3− + K + + Cl − → AgCl ↓ + K + + NO3− Silver electrode is connected with the calomel electrode and silver nitrate (AgNO3) placed in the micro burette and potassium chloride (KCl) in the beaker and emf of the cell is measured and plotted against the volume of silver nitrate added. The steep rise in the curve shows the end point of the titration same as in acid-base titration curve.

5.13 Electrochemical Sensors Electrochemical sensors are devices which are used to measure electrical parameters such as potential difference, current, conductance etc., of the sample under analysis. The sensor which is measure the potential difference is called potentiometric sensor and which measure current is called amperometric sensor. Electrochemical sensors produces an electrical signal which is related to the sample under study. Biological processes such as analysis of glucose in blood and urea are analysed by potentiometric or amperometric sensor.

5.13.1 Potentiometric Sensor A potentiometric sensor is a type of chemical sensor which measure potential difference of an electrode when there is no current flow. Principle Potentiometric sensor is used to determine analytical concentration of gas or solution. Working e lectrode and the reference electrode gives potential difference which is measured by potentiometer. In the potentiometric sensor the ion-selective electrode (ISE) is coupled with the reference electrode to complete electrical circuit and the sensor measured the potential difference between two electrodes is shown figure. Ag. Ag Cl

Ag. Ag Cl

Internal solution Glass membrane Ion selective

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Liquid junction Reference

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Glass electrode is used to measure pH of the solution taken as ion-selective electrode and connected with reference electrode.

5.13.2 Analysis of Glucose in Blood For analysis of glucose in blood glucose sensor which is a potentiometric sensor is used. Glucose is converted into ions, which is detected by ion-selective electrode (ISE). Glucose is oxidised into gluconic acid which further undergoes decomposition and gives H+ ions which are detected by pH electrode. The reaction can be written as oxidase Glucose + O2 Glucose  → Gluconic acid + H 2 O2

Formed H2O2 undergoes reaction at the electrode as shown below H 2 O2 electrode  → 2H + + O2 + 2e − H+ ions is measured by pH meter i.e, glass electrode and a potential difference is set-up between glass electrode and reference electrode which is sense by potentiometric sensor which analyse the glucose level in blood.

5.13.3 Analysis of Urea Analysis of urea in serum or urine sample is very common. For this analysis, enzymatic p otentiometric sensor or urea bio sensors is used. Urea concentration is determined during enzymatic reaction of urea with urease which release NH +4 ions and HCO3− ions. By using ammonium ion-selective electrode analyse the ammonium ions + ( NH 4 ) concentration. NH 2 CONH 2 + 2H 2 O + H + Urease → 2 NH 4+ + HCO3− Electrode is modified with a gel containing the urease enzyme. The signal is determined by potentiometric bio sensor which sense the presence of urea in different sample.

5.14 VOLTAMMETRY Amperometry is an electrochemical technique in which a current is measured as a function of an independent variable, that is, time or electrode potential. Voltammetry is a sub-class of amperometry in which current is measured by varying the potential applied to the electrode. Polarography is a subclass of voltammetry that uses a dropping mercury electrode as the working electrode. Coulometry uses applied current or potential to completely convert an analyte from one oxidation state to another. In these processes, the total current passed is measured directly or indirectly to determine the number of electrons passed. Potentiometry measures the potential of a solution between two electrodes. Here, one electrode is used as a reference electrode; it has the constant potential, and the other is used as an indicator electrode, whose potential changes depend on the sample. Electroanalytical methods that depend on the measurement of current as a function of applied potential are called voltammetric methods. Voltammetry comprises a group of electroanalytical methods that are based upon the potential current behaviour of a polarizable electrode in the solution being analysed.

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5.46 Engineering Chemistry In voltammetry, a measured small potential is impressed across a pair of electrodes one of which is a non-polarizable reference electrode and the other a polarizable inert electrode. The current, which flows, depends upon the composition of the solution. In other words, voltammetry and voltammetric analysis are concerned with the study of current voltage relation at a micro electrode called working electrode. In order to ensure polarization of the electrode, its dimensions generally are made small. Therefore, electrode may be of some inert metal, such as platinum or gold. A three-electrode cell is, however, preferred in general voltammetry. The third electrode can be a simple wire of platinum or silver or mercury pool. The reference electrode may be of any convenient form since it does not carry current (Figure 5.21). Several voltammetric techniques such as Linear sweep voltammetry, Staircase voltammetry, Square wave voltammetry, Cyclic voltammetry, Anodic stripping voltammetry, Cathodic stripping voltammetry, Adsorptive stripping voltammetry, Alternating current voltammetry, Polarography,Rotated electrode voltammetry, Normal pulse voltammetry, Differential pulse voltammetry, and Chronoamperometry play their own importance roles. −

B C

Current meter Working electrode

Digital volt meter

Counter electrode Reference electrode

Figure 5.21 Voltammetric instrument

5.14.1 Linear Sweep Voltammetry (LSV) The effectiveness of this technique is based on its capability for observing the redox behaviour rapidly over a long potential range. Since a rapid linear sweep of the potential is employed, the technique is commonly termed as linear sweep voltammetry (LSV) or stationary electrode voltammetry (SEV). Only stationary or quasi-stationary electrode like dropping mercury electrode is employed as the indicator electrode in an unstirred solution. During the early period, such fast-scan rates could only be monitored over a cathode-ray oscilloscope; therefore, the LSV was then known as cathode-ray polarography. Now-a-days, fast-scan rate X-Y recorders can be used during such experiments. LSV is usually applied at a mercury drop or at any solid stationary electrode, wherein the potential scan rate can be moderate to fast. When a DME is used, the entire potential range is covered on a single drop. Since the sweep rate is fast as compared to drop-time, the recording is made during the end period of the drop’s life. This is done to minimize the condenser or charging current component. It also ensures that the growth of the mercury drop during the sweep period remains negligible.

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In the single-sweep method, the potential of the working electrode increases linearly to a fixed value. It may then fall instantaneously to its starting value. The potential sweep has the appearance of a saw-tooth. The entire i-E curve is recorded during the linear rise of the applied potential. Due to fast potential sweep, there occurs a depletion of the depolarizer around the electrode surface and a peak-shaped i-E curve is recorded. The potential of the peak is characteristic of the depolarizer and its length on the current axis is proportional to concentration of the depolarizer. The following relation holds good for a reversible system. 0.029 E p = E1/ 2 ± n Where Ep is the peak potential, E1/2 is the equivalent de polarographic half-wave potential and n the number of electrons taking part in the electrode reaction. The positive sign holds for the anodic reaction, whereas the negative sign holds for cathodic reaction. Linear sweep voltammetry has been employed both for qualitative and quantitative analyses. It has a reasonably low detection limit, which goes down to 10−5 m. The sensitivity is improved by increasing the scan rate. LSV, being a transient technique, was earlier known as chronoamperometry with potential sweep since the potential axis may be taken as time axis. During periodic polarization (multi-sweep), a saw-tooth voltage is applied with delay. A controlled drop-time is ensured with a DME. Second cycle Switching potential

First cycle

Figure 5.22 A triangular excitation signal applied in cyclic voltammetry

5.14.2 Ferric Fe3+/Fe2+ System Unlike potential step measurements, in LSV measurements, the current response is plotted as a function of voltage rather than time. Fe3+ + e − Fe2 + For the abovementioned system, if an electrolyte solution containing only Fe3+, then the following voltammograms would be seen for a single voltage scan. Current V2 Voltage Time

Figure 5.23

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5.48 Engineering Chemistry The scan begins from left-hand side of the current/voltage plot where no current flows. As the voltage swept further to more reductive values, that is, towards right, a current begins to flow and reaches a peak before dropping. To justify this behaviour, we need to consider the influence of voltage on the equilibrium established at the electrode surface. The rate of electron transfer is fast in comparison to the voltage sweep rate in electrochemical reduction of Fe3+ to Fe2+. An equilibrium is established identical to that predicted by thermodynamics at the electrode surface. Nernst Equation

3+ RT  Fe  E=E + ln nf  Fe2 +  0

Nernst equation can explain the relationship between concentration and voltage or potential difference. Where E = Applied potential difference, E0 = Standard electrode potential. Hence, when the voltage swept from V1 to V2, the equilibrium position shifts from V1 (no conversion) to V2 (full conversion) of the reactant at the electrode surface. The exact form of the voltammogram can be justified by considering the voltage and mass transport effects. When the voltage is initially swept from V1, the equilibrium at the electrode surface begins to alter and the current begins to flow in the following ways: Fe3+ + e − ← Fe2 + Fe3+ + e − Fe2 + Fe3+ + e − Fe2 + Fe3+ + e − Fe2 + Fe3+ + e − → Fe2 + The current rises as the voltage is swept further from its initial value as the equilibrium position is shifted further to the right due to conversion of more reactant. The peak occurs, and at the same point, the diffusion layer has grown sufficiently above the electrode so the flux of reactant to the electrode is not fast enough to satisfy Nernst equation. In this situation, the current begins to drop just as it is in the potential step measurement. The drop in current follows the same behaviour, which can be explained by Cottrell equation. The above voltammogram recorded at a single scan rate. If the scan rate alters the current, then the response also changes. Figure 5.24 shows a series of linear sweep voltammograms recorded at different scan rates for an electrolyte solution containing only Fe3+. Current V2 Increasing scan rate

Time V1

Epc

V2 Voltage

Figure 5.24

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Each curve has same form except total current. Here, total current increases with increasing scan rate. This again can be justified by considering the size of the diffusion layer and the time taken to record the scan. If the scan rate decreases, then LSV voltammogram will take longer time to record. The size of the diffusion layer above the electrode surface will be different depending on voltage scan rate. The diffusion layer will grew much further from the electrode in slow voltage scan when compared to fast scan. Hence, the flux to the electrode surface is smaller at slow scan rate than fast scan rate. “Current is proportional to the flux towards the electrode and the magnitude of the current will be lower at slow scan and higher at high scan rate”. “The position of the current maximum peak occurs at the same voltage; this is the important characteristics of electrode reaction which have rapid electron transfer kinetics and also often referred to as reversible electron transfer reaction”. If the electron transfer processes were slow relative to the voltage scan rate, then the reactions are known as quasi-reversible or irreversible electron transfer reactions. Figure 5.25 represents a series of voltammograms recorded at a single-voltage sweep rate for different reduction rate constants. Current

Decreasing rate constants

Voltage

Figure 5.25 In this kind of situations, the voltage applied will not result in the generation of concentrations at the electrode surface according to the Nernst equation. As the kinetics of the reaction is slow, the equilibria are not established rapidly according to voltage scan rate. In this kind of situation, the overall voltammogram recorded is similar to the graph shown in Figure 5.25, but unlike reversible reaction, the position of the current maximum shifts depending on the reduction rate constant and the voltage scan rate. This is due to the reason that current takes more time to respond to the applied voltage than the reversible case.

5.14.3 Cyclic Voltammetry Cyclic voltammetry is a very versatile and useful technique and is ideal for studying the mechanism of redox reactions of diffusion coefficients and half-cell potentials. Actually, it is a logical extension of LSV. In cyclic voltammetry, a triangular potential wave is impressed on a solid indicator electrode. Therefore, both forward and reverse reactions are monitored, and complete current-potential characteristics of a redox system are supplied. On repeating the triangular potential excitation signal during cyclic voltammetry, the potential of the electrode charges back and forth between two fixed values. These are known as the switching potentials. During a positive scan, the potential becomes increasingly positive, whereas the reverse is known as a negative scan. For recording the cyclic voltammogram, the resulting current is plotted during the positive and negative potential scans.

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5.50 Engineering Chemistry The cyclic voltammogram of a reversible system is shown in Figure 5.26. Here, reduction occurs during the negative scan and the reduced product is oxidized back during the positive scan. The cathodic and anodic peak heights appear equal in a reversible system. The nature and shape of a cyclic voltammetry curve is altered, if the reaction is accompanied by adsorption. Cyclic voltammetry is, thus, useful in diagnosing the mechanism of the electrode reactions. It also enables one to detect unstable intermediates of electrode reaction. Current Y

ipc

X V1

V2 Voltage ipa

Epa Epc

Figure 5.26 A typical cyclic voltammogram In a reversible system, the difference between anodic and cathodic peak potentials is expressed by the following expression at 25°C. E p = ( E p )anodic − ( E p )cathodic =

0.059 n

A reversible redox couple can be identified from its cyclic voltammogram by measuring the potential difference between the two peaks. The mean between the two peak potentials will correspond to the formal electrode potential (E°) of the redox couple. When the potential scan rate (V) is increased, both (ip)anodic and (ip)cathodic increase in proportion to V1/2 . A plot of (ip)a and (ip)c versus V1/2 is linear for a reversible system. The ratio of (ip)a/(ip)c remains unity and independent of the scan rate for a reversible couple with no kinetic complications. This ratio is influenced by coupled chemical reactions and such behavioural study makes cyclic voltammetry a very powerful tool for studying electrode reactions and their mechanisms. The spacing between the two peak potentials is larger with irreversible systems as compared to reversible systems. With increasing irreversibility, the peaks get rounder, and finally, one of them disappears. The difference between the two peak potentials (∆Ep) is used to calculate the heterogeneous charge transfer rate constant after changes occur in the shape of CV on repeating. The greatest utility of cyclic voltammetry is its ability to generate a species during a potential scan and then study its fate in the subsequent scan(s).

5.14.4 Applications of Voltammetry Modern voltammetric methods continue to be potent tools used by analytical, inorganic, physical and biological chemists for fundamental studies of oxidation and reduction processes in various media, adsorption processes on surfaces and electron transfer mechanisms at chemically modified electrode surfaces.

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Several types of voltammetry in current use are as follows: (i) Voltammetric sensors have lot of importance in the determination of specific species that are of interest in industry and research; these devices are sometimes called electrodes but are in fact complete voltammetric cells and are better referred to as sensors. (ii) Stripping voltammetry is a significant trace analytical method, particularly for the determination of metals in the environment. Anodic stripping voltammetryis a quantitative, analytical method for trace analysis of metal cations. The analyte is deposited (electroplated) onto the working electrode during a deposition step and then oxidized during the stripping step. The current is measured during the stripping step. Cathodic stripping voltammetryis a quantitative, analytical method for trace analysis of anions. A positive potential is applied, oxidizing the mercury electrode and forming insoluble precipitates of the anions. A negative potential then reduces (strips) the deposited film into solution. (iii) Differential pulse polarography and rapid scan voltammetry are important for the determination of species of pharmaceutical interest. (iv) Polarographyis a sub-class of voltammetry where the working electrode is adropping mercury electrode(DME), useful for its wide cathodic range and renewable surface. (v) Voltammetry and other electrochemical detectors are frequently employed in high-performance liquid chromatography and capillary electrophoresis. (vi) Amperometric techniques are widely used in sensor technology, monitoring titrations and reactions of biological interest. (vii) Adsorptive stripping voltammetry is a quantitative, analytical method for trace analysis. The analyte is deposited simply by adsorption on the electrode surface (i.e., no electrolysis) and then electrolyzed to give the analytical signal. Chemically modified electrodes are often used. (viii) Rotated electrode voltammetry is ahydrodynamic techniquein which the workingelectrode, usually arotating disk electrode(RDE) orrotating ring-disk electrode(RRDE), is rotated at a very high rate. This technique is useful for studying thekineticsandelectrochemical reaction mechanismfor ahalf-reaction.

5.15 BATTERIES A battery is a device in which numbers of electrochemical cells are connected in series. It converts chemical energy into electrical energy at a constant voltage. Batteries are generally used at a commercial level.

5.15.1 Advantages of Batteries (i) (ii) (iii) (iv)

In batteries, there is no need to be connected to an electrical system. Batteries act as a portable source of energy. Batteries can be easily replaced. Different types of batteries have their specific applications e.g. miniature batteries (primary cell) is generally used in electric watches, calculator, medical devices and secondary cells are generally used in portable equipments like portable radio and TV, mobile phones, lap–top, computersetc.

5.15.2 Disadvantages of Batteries (i) Batteries can be used for only a limited time, even rechargeable batteries can be recharged a certain number of times.

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5.52 Engineering Chemistry (ii) Some batteries are dangerous and can lead to fire, explosion, chemical pollution etc. (iii) Some types of batteries need to be maintained and checked periodically. Requirements should be Possessed by the Batteries (i) (ii) (iii) (iv)

It should have the compactness, lightness and ruggedness for portability. Long life cycle is required. Operating voltage is required i.e., voltage should not drop much during use. It should be stable with time, temperature, vibration shock etc.

Classification of Various Commercial Cells (i) Primary cells (ii) Secondary cells (iii) Fuel cells (i) Primary cells: Primary cells are those in which redox reaction occurs only once and cell becomes dead after some time. These are non-rechargeable and cell becomes dead after some time. For example: dry cell, mercury cell and lithium cell etc. (ii) Secondary cells: Secondary cells are those which can be recharged by passing an electric current through them and hence can be used over and again. For example: Lead storage cell, Ni – Cd storage cell etc. (iii) Fuel cells: Fuel cells are those in which the energy produced from the combustion of fuel such as H2, O2, CH4 etc. is directly converted into electrical energy. A Brief Description of Example of Each Type of Cell (i) Primary Batteries (a) Dry cell: Dry cell is a compact form of the Leclanche cell. This is also known as carbon – zinc cells. It consists of a cylindrical zinc container which acts as the anode. A graphite rod placed in the center acts as the cathode. Dry cell does not use any liquid that’s why it is called as dry cell. It consists of a mixture of NH4Cl, ZnCl2 and MnO2 made into a paste by the addition of starch which acts as electrolyte. The zinc container is covered with cardboard to protect it from the atmosphere as shown in Figure 5.27. Metal cap

(+)

Pitch seal

Zinc container (Anode) Graphite (Cathode) Card board cover

MnO2 + C Paste of NH4Cl + ZnCl2

Figure 5.27 The dry cell

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The reactions of the cell are: At anode: Zn(s) → Zn 2+ (aq) + 2e− At cathode: 2MnO2 (s) + 2 NH 4+ (aq ) + 2e − → Mn 2 O3 (s) + 2 NH 3 (g) + H 2 O Overall reaction: Zn(s) + 2 NH +4 (aq ) + 2MnO2 (s) → Zn 2 + (aq ) + Mn 2 O3 (s) + 2 NH 3 (aq ) + H 2 O(l) The NH3 formed is combined immediately with the Zn 2+ ions and Cl− ions to form the complex salt [Zn (NH3)2Cl2] Zn2+ + 2NH3 + 2Cl− → [Zn (NH3)2]Cl2 The dry cell gives a voltage of about 1.5 V. The dry cell is generally used in flash lights, calculators, toys etc. (b) Lithium cells: Lithium cells are primary (disposable) batteries. Lithium cells have lithium metal as anode comprising many type of cathodes and electrolytes. Lithium cells are safer, less expensive, and non-toxic and meet the needs of present as well as future generation. Due to its high electrode potential, it can produce voltages from 1.5 V to about 3.7 volt, which is twice the voltage of an ordinary battery. Lithium cells are generally used in portable consumer electronic devices like radios, clocks, MP3 players, hearing aids, heart pacemakers etc. Lithium cells can be classified into three categories: (1) Lithium cells with solid cathodes (2) Lithium cells with liquid cathodes (3) Lithium cells with solid electrolyte (1) Lithium cells with solid cathodes: These cells use solid cathode materials such as MnO2, CuO, V2O5 and carbon monofluoride (CF)n. They cannot be discharged as rapidly as liquid cathode cells. Most commonly lithium batteries in use are of the Li|MnO2 type. Due to its low discharge characteristic it is generally suitable for memory backup, watches, calculators, cameras, etc. The Li|MnO2 cell gives a voltage of about 3.2 V and it is capable to perform at low or high discharge rates on pulse and perform at wider range of temperature of the cell. It has a shelf life of around 5–7 years. Anode: Li metal act as Anode. Cathode: Mixture of heat – treated electrolytic manages dioxide (MnO2) and conducting agents. Electrolyte: Mixture of propylene carbonate and 1, 2-dimethoxy ethane. Cell reaction At Anode: Li → Li+ + e− IV

III

At cathode:

M nO2 + Li + + e − → MnO2 ( Li + )

Overall reaction:

Li + M nO2 → MnO2 ( Li + )

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5.54 Engineering Chemistry Oxidation of lithium metal at the anode takes place which produce positively charged lithium ions (Li+) and electrons (e−). The Li+ ions go into the solutions and diffuse through electrolyte and electrons (e−) reach at the cathode where MnO2 is reduced from tetravalent to trivalent state. (2) Lithium cells with liquid cathodes: In these cells, the cathode material is reduced during discharging is present in the liquid form. Examples are • SO2 dissolved in a solution of Lithium bromide and acetonitrile • Liquid thionyl Chloride (SOCl2) as a solvent with lithium aluminium chloride as solute. It is incompatible with standard batteries, that why they are mostly used for military applications such as munitions, transceivers and surveillance equipments. This type of cells gives a voltage of about 2.8 V to 3.5 V for different systems. It has a shelf life about 10 years at normal and cooler temperature. (i) In Li-SO2 cell Anode: Lithium metal Cathode: SO2 Electrolyte: Lithium bromide and small amount of acetonitrile There is a liquid cathode; it forms a protective layer at the interface of the lithium and SO2. To maintain the SO2 in a liquid form, Li–SO2 cells internal pressure is 3 bars at +20 °C temperature and 14 bars at +70 °C temperature. Overall reaction 2Li + 2SO2 → Li2S2O4 (ii) In Li–SOCl2 cell Anode: Lithium metal Cathode: SOCl2 Electrolyte: LiAlCl4 (lithium terachloro aluminate) In this, porous carbon material acts as a cathode current collector which receives electrons from the external circuit. It is not sold in the consumer market but having more application in industrial as well as military level. Thionyl chloride is a corrosive liquid and reacts with lithium to produce LiCl, S and SO2. LiCl is precipitated on carbon electrode and SO2 and S are soluble in electrolyte, at lower depth of discharge. 4Li + 2SOCl2 → 4LiCl + SO2 + S Due to incomplete protective layer at surface of lithium, additives are used. It provides a voltage of around 3.5 V. It has poor shelf life. (3) Lithium cells with solid electrolyte: In such type of cell, solid such as lithium iodide (LiI) ﹛which are electronic insulators but ionic conductors﹜ can be used as the electrolyte in solid electrolyte batteries. Such batteries have extremely long shelf life at low drain currents or even at high temperature. Such types of cells are generally used in heart pacemakers, and for preserving volatile computer memory. Anode: Li metal Cathode: Poly-2-vinyl pyridine (p2vp) Electrolyte: Solid LiI (In situ)/ I2 (iodine)

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There is a solid electrolyte and it provides a voltage of around 2.8 V. LiI is formed in situ by direct reactions of the electrodes. Overall reaction 2Li + p2Vp + nI2 → 2LiI + p2vp + (n − 1) I2 It is suitable only for low-current application and its voltage decreases with degree of discharge due to precipitation of lithium iodide. It has very low self–discharge voltage. (ii) Secondary batteries: (a) Lead-acid cells: or (Lead storage cell) This type of cells is called as storage battery or a storage cell. A storage cell is that which can work both as voltaic cell as well as electrical cell. During its operation it works as a voltaic cell i.e., it supplies electrical energy due to a chemical energy and it becomes discharged. Then during its charging, it works as an electrolytic cell i.e., chemical is produced by supplying of electrical energy. It is also called as lead storage battery. It consist of Anode: Lead (Pb) act as anode (or grid of lead filled with spongy lead) Cathode: Lead dioxide (PbO2) (or grid of lead filled with PbO2) Electrolyte: Dilute sulphuric acid (H2SO4) (35% by weight) A lead storage battery consists of 6 cells each producing 2V output. To increase the current output of each cell, the cathode plates are joined together and the anode plates are also joined together and connected in series, we get an output of 12V as shown in Figure 5.28. Anode (+) (−) Cathode

Lead (Pb)

Dil H2SO4

Lead dioxide (PbO2)

Figure 5.28 Lead storage battery Electrode reactions during discharge At anode: Lead (Pb) undergoes oxidation Pb → Pb2+ + 2e− The Pb2+ ions combine with SO42- (H2SO4) ions to produce PbSO4 Pb2+ + SO42- → PbSO4↓ At cathode: Lead Oxide (PbO2) undergoes reduction due to flow of electrons from anode to cathode PbO2 + 4H+ + 2e− → Pb2+ + 2H2O

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5.56 Engineering Chemistry and the Pb2+ ions combines with SO42− ions to produce PbSO4 Pb2+ + SO42- → PbSO4↓ Overall reaction Pb + PbO2 + 4 H + + 2SO24 − → 2PbSO4 + 2H 2 O + Energy Electrode reaction during charging The lead storage battery is rechargeable. Now the cell is operated like an electrolytic cell. The following reactions occur during charging: At Anode: PbSO4 + 2e − → Pb + SO24 − At Cathode: PbSO4 + 2H 2 O → PbO2 + SO24 − + 4 H + + 2e − Overall reaction: 2PbSO 4 + 2H 2 O → Pb + PbO2 + 4 H + + 2SO24 − The PbSO4 formed during discharge is a solid and sticks to the electrodes. So, it is in position to gain or receive the electrons during electrolysis. Such type of cell is generally used for electrical vehicles, Automobiles, railway, laboratories, hospitals, power stations, in telephone exchange, UPS system etc. (b) Ni-Cd Storage cell (or NiCad cell): This is also rechargeable cell which is generally used in calculators. It has a longer life than lead storage cell. It consist of Anode: Cadmium (Cd) electrode Cathode: Nickel (III) oxide-hydroxide [NiO (OH)] Electrolyte: Alkaline electrolyte (KOH) Normally Ni-Cd cells have a potential of 1.2V and by using six cells, a voltage of 7.2V can be produced. Electrode reactions during discharge At Anode: Cd + 2OH− → Cd (OH)2 + 2e− At Cathode: 2NiO (OH) + 2H2O + 2e− → 2Ni (OH)2 + 2OH− Overall Reaction: Cd + 2NiO (OH) + 2H2O → Cd(OH)2 + 2Ni(OH)2 During recharge, the reactions go from right to left. The alkaline electrolyte (KOH) is not consumed in this reaction. Such types of cells are used in cordless and wireless telephones, emergency lighting, remote controlled electric model airplanes, boats, and cars etc.

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(c) Lithium ion cells: Lithium ion batteries (sometimes abbreviate as Li-ion batteries) are a type of rechargeable battery in which the cathode (positive electrode) contains lithium and the anode (negative electrode) is generally made of a type of porous carbon. During discharging, the current flows within the battery from the anode to the cathode; the internal process is the movement of Li+ ions from anode to the cathode, through the non-aqueous electrolyte. During charging, an external power source, the current to pass in the reverse direction. The positive terminal of the charging circuit is connected to the cathode of the battery and negative terminal is connected to the anode. Anodes: Hard carbon (LiC6), Graphite (LiC6) Cathodes: LiCoO2, LiMn 2O4, LiNiO2 Electrolytes: Liquid (non-aqueous) electrolyte which consists of lithium salts such as LiPF6, LiBF4 in an organic solvent, such as ethylene carbonate. The cathode half reaction   → Li1− x CoO2 + xLi + + xe − LiCoO2 ←  The Anode half reaction   → Li x C6 xLi + + xe − + 6C ←  In Lithium-ion battery the lithium ions are transported to and from the cathode or anode, with the transition metal, Co in Li1-xCoO2 being oxidized from Co3+ to Co4+ during charging, and reduced from Co4+ to Co3+ during discharge. Applications: Used in cell phones, laptops, electric equipments. Lithium-ion batteries are common in portable consumer electronics because of high energy-to-weight ratios, lack of memory effect, and slow-discharge when not in use. Lithium-ion batteries are not to be confused with lithium batteries; the difference is that lithium batteries, containing metallic lithium acts as primary batteries, while lithium-ion batteries are secondary batteries, containing an intercalation anode material. (iii) Fuel cells: Fuel cell is a device which converts the energy produced during the combustion of fuels directly into electrical energy. The process in a fuel cell is: Fuel + oxygen → oxidation products + Electricity (oxidant)

Fundamental principles of fuel cell and electrochemical cells are the same, but only difference between them is that in fuel cell, chemical energy is provided by a fuel and oxidant is stored outside the cell in which that reactions take place. Fuel cells have the most important characteristics: (a) High efficiency (b) Low emission levels (c) Low noise levels (d) Fuel cells are free from vibration, heat transfer and thermal pollution

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5.58 Engineering Chemistry Examples of Fuel Cells (i) Hydrogen-oxygen fuel cell: A common type of fuel cell based on the combustion of hydrogen gas: 2H2(g) + O2(g) → 2H2O(l) This is known as Hydrogen–oxygen fuel cell. The design of H2–O2 fuel cell is as shown in Figure 5.29. Load 4e− Anode −

+ Cathode Gas chamber

Porous electrodes Ion-exchange membrane

Fuel (Hydrogen)

Oxidizer (Oxygen)

4e− 2H2

4e−

4H+

O2 2H2O

H2O

Figure 5.29 H2–O2 fuel cell It consists of two electrodes made of porous graphite impregnated with a catalyst (platinum, silver or a metal oxide). The electrodes are placed in an aqueous solution of KOH or NaOH. Through the anode, H2 gas is bubbled and through the cathode, oxygen gas is bubbled under pressure of about 50 atm. The gases diffuse into the electrode pores and so does the electrolyte solution. The half-cell reactions which occur at the electrodes are as follows: At Anode: H2(g) + 2OH− (aq) → 2H2O + 2e− (Oxidation half reaction) At Cathode: O2(g) + 2H2O + 4e− → 4OH−

(Reduction half reaction)

Overall fuel cell reaction: 2H2(g) + O2(g) → 2H2O(l) The standard emf of the H2–O2 cell is E° = E°(cathode) - E°(anode)

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= E°(O2/OH-) -E°(H2O/H2) = 0.401 - (-0.828) E° = 1.229 V In actual practice, the emf of the cell is 0.8 to 1.0 V. A fuel cell battery or fuel battery comprise of an arrangement of such cells in parallel or series (modules/stack of cells). Uses of H2-O2 Fuel Cells (i) H2-O2 fuel cell is generally used in space craft because of their high efficiency, lightness and product water is a source of fresh water for the astronauts. (ii) It is also used in submarines or other military vehicles. (iii) This cell is used as the primary source of electrical energy on the Apollo moon flights. Approximately 200 kg of fuel is sufficient for 11 days in space. Types of H2-O2 Fuel Cell On the basis of the type of electrolyte used in fuel cell, they are of five types: Fuel cell

Electrolyte used

(i) Alkaline fuel cell (AFC) (ii) Molten carbonate fuel cells (MCFC)

Aqueous KOH (30–40%) Mixture of Na2CO3 + WO3

(iii) Phosphoric acid fuel cells (PAFC)

Phosphoric acid (H3PO4)

Mixture of yttrium dioxide (Y2O3) + Zirconium dioxide (ZrO2) (v) Proton exchange membrane fuel cells (PEMFC) A H-form (cation exchange resin membrane) is used in place of a fluid electrolyte.

(iv) Solid oxide fuel cells (SOFC)

Fuel Cells and their Characteristic Features Characteristic features

AFC

MCFC

PAFC

SOFC

PEMFC

Primary fuel Electrodes Electrolytes

H2 Carbon Aqueous KOH (30–40%)

H2, CO, CH4 Stainless steel Molten sodium carbonate

H2 Graphite Phosphoric acid

H2, CO Ceramic Yttriumstabilized ZrO2

H2 Carbon Polymer membrane

Catalyst Charge carrier Operating temperature Power density (kW/m3) Combined cycle fuel cell efficiency Major applications

Pt OH− 50–200 °C

Ni CO32600–700 °C

Pt H+ 150–220 °C

Perovskite O2700–1000 °C

Pt H+ 50–100 °C

1.5–2.6

0.8–1.9

0.1–1.5

4.8–6.5

50-60%

50-70%

55%

55-65%

50-60%

Stationary power

Auxiliary power Stationary and in vehicles automotive power

Space vehicles and Stationary drinking water power

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5.60 Engineering Chemistry (ii) Methanol-oxygen fuel cell: The Half cell reactions for such cells are: At anode: CH3OH(l) + H2O(l) → CO2(g) + 6H+(aq) + 6e- (O.H.R) 3 O (g) + 6H + (aq ) + 6e − → 3H 2 O(l) (R.H.R) At cathode: 2 2 3 CH 3 OH(l) + O2 (g) → CO2 (g) + 2H 2 O(l) Overall reaction: 2 Advantages of Fuel Cell (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

The energy conversion (chemical into electrical) is very high by fuel cells (75-82.8%). H2-O2 fuel cell produces H2O, which is used as drinking purpose by astronauts. Noise and thermal pollution are low. Such type of cells never becomes dead, because of continuous supply of fuel. Modular and other parts of fuel cells are exchangeable. Fuels cells having low maintenance cost. Fuels cells saves fossil fuels. The regenerative H2–O2 fuel cell is an energy storage system for space application, submarines and other military vehicles.

Limitations of Fuel Cells (i) (ii) (iii) (iv) (v) (vi)

High initial cost. Large weight and volume of gas-fuel storage system. High cost of pure hydrogen. Liquifaction of hydrogen requires 30% of the stored energy. Life-times of such type of cells are not accurately known. Most alkaline cells suffer from CO2 degradation and hence CO2 should be removed from the fuels and the air.

5.16 Review Questions 5.16.1 Fill in the Blanks 1. The electrolyte whose solution conducts electricity to a small extent is called a _______ [Ans.: Weak electrolyte] 2. The electrolytic conductance _______ with temperature. [Ans.: increases] 3. The unit of specific conductance is _______ [Ans.: ohm−1 cm−1] 4. Specific conductance is the conductance of _______ of the solution [Ans.: 1 cm3] 5. The equivalent conductance _______ with dilution. [Ans.: Increase] 6. The L°For a weak electrolyte is calculated by using _______ law. [Ans.: Kohlrausch]

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7. An electrochemical cell converts _______ energy into _______ energy. [Ans.: Chemical, electrical] 8. The electrode used in the fuel cell is _______. [Ans.: Concentrated KOH solution] 9. The effect of concentration on the electrode potential is studied by _______ equation. [Ans.: Nernst] 10. Arrangement of electrodes in order of reducing potential is known as _______. [Ans.: Electrochemical series] 11. Calmol electrode having _______ solution an electrolyte [Ans.: saturated KCl] 12. A device which convert the energy of fuel direct into electrical is called as _______ cell. [Ans.: fuel] 13. The substance which conducts electricity without any decomposition is called as _______ conductor. [Ans.: electronic] 14. A unit of molar conductivity is _______. [Ans.: ohm−1 cm2 mol−1] ΛC 15. Degree of dissociation (α) = . [Ans.: L0] 16. The tendency of electrode to loss or gain electrode is called as _______ . [Ans.: Electrode potential] 17. Calomel electrode act as _______ electrode. [Ans.: reference] 18. Electrode potential of saturated calomel electrode is _______ [Ans.: 0.2422 V] 19. When two like electrodes at different concentrations are dipped in some solution of the electrolyte is called as _______ concentration cell. [Ans.: Electrode] 20. _______ cells are not reversible in nature. [Ans.: Primary] 21. Lithium ion cell is an example of _______ cell. [Ans.: Secondary] 22. In Lead-acid cell _______ is act as electrolyte. [Ans.: Dilute H2SO4]

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5.62 Engineering Chemistry

5.16.2 Multiple-choice Questions 1. The effect of temperature increases on concentration is as follows: (a) Metallic conduction increases, electrolytic conduction decreases (b) electrolytic conduction increases, metallic conduction decreases (c) Both metallic and electrolytic conduction decreases (d) Both metallic and electrolytic conduction increases [Ans.: b] 2. The units of equivalent conductance are: (a) Ohm−1 (c) Ohm−1 cm2

(b) ohm−1 cm−2 (d) mho cm 2

[Ans.: c] 3. The cell constant is l a (a) (b) a l (c) a × l

(d)

[Ans.: a]

K R

4. Effect of dilution on conduction is as follows: (a) Specific conductance increases, molar conductance decreases (b) Specific conductance decreases, molar conductance increases (c) Both increases with dilution (d) Both decreases with dilution [Ans.: b] 5. The units of the cell constant is (a) cm−1 (c) cm3

(b) cm 2 (d) cm−2

[Ans.: a] 6. The potential of a single electrode is a half cell is called as (a) Reduction potential (b) Half-wave potential (c) Single electrode potential (d) Cell potential [Ans.: a] 7. Which of the following constitutes Daniel Cell? (a) Zn − Ag cell (b) Cu − Ag cell (c) Zn − Cu cell (d) none of these [Ans.: c]

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8. EMF of a cell in terms of reduction potential of its left and right electrode (a) E = Eleft + Eright (c) E = Eright − Eleft

(b) E = Eleft − Eright (d) None of these

[Ans.: c] 9. Which of the following is a secondary cell (a) Dry cell (c) Ni – Cd cell

(b) Mercury cell (d) H2 – O2 cell

[Ans.: c] 10. An electrochemical cell stops working after some time because (a) One of the electrodes is eaten away (b) electrode potentials of both electrodes becomes equal in magnitude (c) electrode potentials of both the electrodes go on decreasing (d) electrode potentials of both the electrodes go on increasing [Ans.: b] 11. The standard EMF (E°) for the cell reaction Zn + Cu 2+ → Zn2+ + Cu is 1.1 volt at 25 °C. The EMF(E) of the cell reaction when 0.1M Cu 2+ and 0.1 M Zn2+ solutions are used, at 25 °C is (a) 1.10 V (c) −1.10 V

(b) 0.10 V (d) −0.110 V

[Ans.: a] 12. In an electrochemical cell (a) Potential energy decreases (b) Kinetic energy decreases (c) Potential energy changes into electrical energy (d) Chemical energy changes into electrical energy [Ans.: d] 13. As lead storage battery is charged (a) lead dioxide dissolves (b) sulphuric acid is regenerated (c) lead electrode becomes coated with lead sulphate (d) the concentration of sulphuric acid decreases [Ans.: b] 14. In an electrochemical series electrodes are arranged in the (a) Increasing order (downwards) of standard reduction potential (b) Decreasing order of standard reduction potential

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5.64 Engineering Chemistry (c) Increasing order of standard oxidation potential (d) Increasing order of equivalent weight [Ans.: a] 15. Electrode potential of standard calomel electrode is (a) 0.2422 V

(b) 0.2400 V

(d) 0.3335 V

[Ans.: a] 16. Silver – Silver chloride electrode is a type of (a) Redox electrode

(b) Metal – Metal Ion electrodes

(d) Gas – ion electrode

[Ans.: b] 17. Calomel electrode consist of calomel with a solution of (a) Saturated NaCl

(b) Saturated Ca (OH)2

(d) Saturated AgCl

[Ans.: c] 18. Which electrode is used for pH measurement (a) Silver electrode (b) Redox electrode (c) Glass electrode (d) Calomel electrode [Ans.: c] 19. Lechanche cell is an example of (a) Primary cell (c) Tertiary cell [Ans.: a]

(b) Secondary cell (d) Fuel cell

20. A fuel cell converts (a) Chemical energy into electrical energy (b) Chemical energy into potential energy (c) Chemical energy into heat (d) Chemical energy into pressure [Ans.: a] 21. Which is produced during H2−O2 fuel cell (a) CH3OH (c) H2O2 [Ans.: b]

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(b) H2O (d) H3O+

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22. The cathode of Ni – Cd cell is made up from (a) NiOH

(b) Ni (OH)2

(d) NiO (OH)

23. During Charging of lead – acid cell, concentration of H2SO4 (a) Increases (b) Decreases (c) Remains unchanged (d) First increases then decreases [Ans.: a] 24. In potentiometric titration, graph is plotted between variations of (a) Electrode potential with temperature of titrant (b) Electrode potential with pressure of titrant (c) Electrode potential with volume of titrant (d) Electrode potential with concentration of titrant [Ans.: c] 25. At equilibrium, EMF of the cell is (a) 0 V (c) Less than 0 V [Ans.: a]

(b) 100 V (d) More than 0 V

26. In glass electrode, the glass membrane undergoes exchange of Na+ ion with (a) Ca2+ (b) Mg2+ (c) H+ (d) NH +4 [Ans.: c] 27. Quinhydrone – electrode consist of hydroquinone (QH 2) and quinone (Q) in the ratio of (a) 1: 2 (b) 1:1 (c) 2:1 (d) 1:15 [Ans.: b] 28. Conductance of an electrode depends upon (a) Number of free ions present in solution (b) Number of free ions present in solvent (c) Concentration of the solution (d) Temperature of the solution [Ans.: a] 29. Four metals A, B, C and D are having their reduction potentials as −3.05, −1.66, −0.40 and −0.80 V respectively. Which one of these will be most reducing agent. (a) A (b) B (c) C (d) D [Ans.: a]

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5.66 Engineering Chemistry 30. Equivalent conductance of NaCl, HCl and C2H5COONa at infinite dilution are 126.45, 426.16 and 91 ohm−1 cm2 respectively. The equivalent conductance of C2H5COOH at infinite dilution is (a) 201.28 ohm−1 cm2 (b) 390.71 ohm−1 cm2 −1 2 (c) 698.28 ohm cm (d) 540.48 ohm−1 cm2 [Ans.: b] 31. Saturated Solution of KNO3 is used to make salt bridge because (a) Velocity of K+ is greater than that of NO3− (b) Velocity of NO3− is greater than that of K+ (c) Velocities of both K+ and NO3− are nearly the same (d) KNO3 is highly soluble in water [Ans.: c] N 32. The specific conductance of solution of KCl at 25 °C is 0.002765 ohm−1. If the resistance of 50 the cell is 400 ohms, then what is the value of cell constant (a) 2 (b) 1.106 (c) 3 (d) 3.2 [Ans.: b] 33. The EMF of the cell Ni | Ni2+ (1.0M) | Au3+(1.0M) | Au Given that  E° = − 0.25V for Ni 2 + | Ni   3+  E° = +1.5V for Au | Au  (a) 1.25 V (b) −1.25 V (c) 1.75 V (d) 2.00 V [Ans.: c] 34. The conductivity of a saturated solution of BaSO4 is 3.06 × 10−6 ohm−1 cm 2 and its equivalent conductance is 1.53 ohm−1 cm2 equiv−1. The Ksp for BaSO4 will be (a) 4 × 10−12 (b) 2.5 × 10−9 (c) 2.5 × 10−13 (d) 4 × 10−6 [Ans.: d] 35. Which of the following does not conduct electricity (a) Molten NaCl (b) Aqueous NaCl (c) Solid NaCl (d) Aqueous NH4Cl [Ans.: c]

5.16.3 Short Answer Questions 1. How can you test whether the given electrolyte is a strong electrolyte or a weak electrolyte? Ans.: If the aqueous solution of the electrolyte conducts electricity to a large extent, it is a strong electrolyte and if to a small extent, it is a weak electrolyte.

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2. Define specific conductivity and mention its units. Ans.: The specific conductance of a solution is defined as the conductance of 1 cm3 of the solution of the electrolyte. Its unit is ohm−1 cm−1 (or Ω−1 cm−1) (or S cm−1) 3. What is the effect of temperature on the electrical conduction of (i) metallic conductor (ii) electrolytic conductor? Ans.: With increase of temperature, the electrical conduction of metals decreases whereas that of electrolyte increases. 4. Define molar conductance and gives its units. Ans.: It is defined as the conductance of all the ions produced by dissolving 1 mole of the electrolyte in V cm3 of the solution. Its units is ohm−1 cm2 mol−1 (or S cm 2 mol−1) 5. Why do electrochemical cells stop working after some time? Ans.: Electrochemical cells produce electrical energy at the cost of chemical energy as spontaneous redox reaction takes place in them. When the redox reaction is completed, the cell stops working. 6. Give the relationship between molar conductivity and specific conductivity. Ans.: 1000 Λm = K × C Λ m = Molar conductivity K = Specific conductivity C = Molar concentration 7. What is the relationship between specific conductance and equivalent conductance. Ans.: Λeq = K × V Λeq = Equivalent conductivity K = Specific conductance V = Volume of solution containing 1g eq of the substance 8. Give the relationship between equivalent and molar conductance. Ans.: Λ m Normality = Λ eq Molarity 9. What is cell constant. Give its units. Ans.: it is the ratio of the distance between the parallel plates of the cell and the area of electrolyte. Cell constant =

l . a

Its unit is cm−1

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5.68 Engineering Chemistry 10. How is cell constant calculated from conductance values. Ans.: cell constant =

spcific conductance observed conductance

11. Why voltmeter cannot be used for the precise measurement of EMF of the galvanic cell. Ans.: Because a part of the cell current is drawn by the voltmeter itself, thereby giving lower value of EMF than the actual one. 12. What flows in the internal circuit of galvanic cells? Ans.: Ions flow in the internal circuits of galvanic cells. 13. What is the EMF of the cell when the cell reaction attains equilibrium? Ans.: At equilibrium, EMF of the cell is zero. 14. Can we use a copper vessel to store 1 M AgNO3 solution, given that E °Cu 2+/Cu = +0.34 V, E°Ag+/Ag = +0.80 V Ans.: As the reduction potential of Ag+/ Ag electrode is higher than that of Cu2+/ Cu electrode. So Cu metal is incapable of displacing silver from silver nitrate solution. Thus, we can use a copper vessel to store 1 M AgNO3 solution. 15. Why electrode potential of zinc is assigned a negative value; whereas that of copper a positive value. Ans.: Because Zn electrode is anodic w.r.t. S.H.E and Cu electrode is cathodic w.r.t. S.H.E. 16. What is the relationship between the standard EMF of the cell and the equilibrium constant of the cell reaction at 298K? 0.0591 log Kc Ans.: E°cell = n Where E °cell = standard EMF of the cell Kc = Equilibrium constant n = number of electrons involved in reactions 17. What is the electrolyte used in a dry cell. Ans.: A paste of NH4Cl, MnO2 and C is used in dry cell. 18. Why the blue colour of the solution gradually fades when CuSO4 solution is electrolyzed using platinum electrodes. Ans.: The blue color is due to the presence of Cu2+ ions. During electrolysis these ions are converted into metallic copper hence the blue color fades. 19. Define Kohlrausch’s law? Ans.: At infinite dilution, equivalent conductance of an electrolyte is equal to the sum of ionic conductances of cations and anions at infinite dilution.

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20. Why D.C current is not used while determining the resistance of an electrolyte? or Why only A.C is used and not D.C source in conductometric estimation? Ans.: If conductometric estimation is carried out by using D.C then the products of electrolysis collect at the electrodes and set up a back emf, Which apparently increases the resistance of the electrolyte. So, such estimations are carried out by using AC and detecting the flow of current. 21. What is the basic reason that a lead storage battery can be recharged? Ans.: PbSO4 is deposited on the electrodes. So, the electrode reactions can be reversed. 22. Out of zinc and tin which are protects iron better even after cracks and why? Ans.: Zinc protects better because oxidation potential of zinc is greater but that of tin is less than that of iron. 23. Why does the equivalent conductivity of a weak electrolyte increase with dilution? Ans.: Because degree of dissociation of weak electrolyte increases with dilution. As a consequence, the total number of ions present per gram equivalent also increases. Hence, equivalent conductivity of weak electrolyte increases with dilution. 24. Why equivalent conductivity at infinite dilution for a weak electrolytes solution cannot be determined experimentally. Ans.: Equivalent conductivity for a weak electrolyte increases steadily with dilution, and the curve between Leq and C is not a straight line. The curve does not meet the equivalent conductivity axis. So, the value of L∞ cannot be obtained by extrapolation. Hence L∞ for a weak electrolyte is determined indirectly by using kohlrausch’s law. 25. Why, with dilution, equivalent conductance increases but specific conductance decreases. Ans.: With increase in dilution, two things happen, namely degree of dissociation increases, but the total volume increases. The number of ions per cm3 decreases, because the effect of increased volume in decreasing the number of ions per cm3 outweighs the minor increase in conductance with increase of dilution. On the other hand, equivalent conductance increases, because the total number of ions per gram equivalent increases with dilution, due to increased degree of ionization. 26. Why a dry cell becomes dead after a long time even if it has not been used? Ans.: This is because the acidic NH4Cl corrodes the zinc container even if it has not been used. 27. Solution of two electrolytes A and B each having a concentration of 0.2 M have conductivities 2 × 10−2 and 4 × 10−4 S cm−1 respectively. Which will after greater resistance to the flow of current and why? Ans.: l 1 l k = c× = × a R a

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5.70 Engineering Chemistry i.e k ∝

1 R

Conductivity is reciprocal to resistance. Hence, B will offer greater resistance. 28. What is the role of ZnCl2 in dry cell? Ans.: ZnCl2 combine with NH3 produces to form the complex salt [Zn(NH3)2Cl2] as otherwise the pressure developed due to NH3 would crack the seal of the cell. 29. Which types of cells are rechargeable? Ans.: Those cells are rechargeable in which the products formed during discharge are deposited on the electrodes and these can be decomposed to give the original substances when electrical energy is supplies. 30. The standard reduction potential values of three metallic cations X, Y, Z are 0.52, -3.03, -1.18 V respectively. What will be the order of reducing power of the corresponding metals? Ans.: The standard oxidation potential (equal and opposite in sign of standard reduction potential) of the metals X, Y, Z will be -0.52, 3.03, 1.18 V respectively. Higher the oxidation potential, more easily metal is oxidized and here greater is the reducing power. Hence, the reducing power will be in the order Y > Z > X. 31. Why a cell stop working after some time Ans.: With time, concentration of the electrolytic solutions change. Hence, their electrode potentials change. When the electrode potentials of two half cells become equal, the cell stops working. 32. Write any two advantage of H2 – O2 fuel cell over ordinary cell. Ans.: (i) They do not cause pollution. (ii) They have high efficiency of 60 – 70%. 33. Write the Nearst’s equation for the electrode reaction: M n + (aq ) + ne − → M(s) Ans.: E M n+ / M = E° M n+ /M +

2.303RT log[ M n + (aq )] nF

34. Define standard hydrogen electrode. Ans.: Standard hydrogen electrode (SHE) or Normal hydrogen electrode (NHE) is a reference electrode which is obtained by dipping platinum foil in 1M HCl solution through which hydrogen gas is passed at 298 K under 1 atm pressure. Its electrode potential is zero. 35. Glass electrode is preferred to quinhydrone electrode in measuring pH of a solution, Give reason. Ans.: Glass electrode is simple, not easily oxidized and attain equilibrium rapidly. It can safely be used upto pH of 10. On the other hand, quinhydrone electrode can be used upto pH of 8 only. It cannot be used in solutions containing redox system. Hence glass electrode is preferred over quinhydrone electrode in pH measurement of a solution.

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36. Alkaline dry cells are considered better than Lechanche cell why? Ans.: Alkaline dry cell lasts longer, because zinc electrode does not corrode easily. 37. What is the purpose of MnO2 in dry cell? Ans.: It acts as an oxidizing agent in dry cell. 38. Write the formula for determining the degree of dissociation of weak electrolyte. Ans.: equivalent conductivity at any concentration equivalent conductivity at infinite dilutioon Λ a = c Λo

Degree of dissociation (a ) =

39. What is the emf of H2 – O2 fuel cell? Ans.: E° = 1.229 V 40. Define fuel cell. Ans.: Fuel cell is a device which converting energy of a fuel directly into electrical energy. 41. What is Lead - acid accumulator? Ans.: A secondary cell consisting as lead electrodes, the positive one covered with PbO2, dipping into H2SO4 solution. Its emf is about 2V. 42. Write major applications of lithium ion cells. Ans.: Lithium ion cell used in cell phones, laptops, electrical equipment’s etc.

5.16.4 Solved Numerical Problems

N KCl solution at 298 K is 0.002765 ohm−1 cm−1 and resistance of a 50 cell containing this solution is 100 ohm. Calculate cell constant.

(i) If specific conductivity of

Solution As we know that Cell constant =

SP. conductivity K = obs. conductance C

Specific conductivity, K = 0.002765 ohm−1 cm−1 Conductance, C =

1 ohm −1 100  l  0.002765 So, cellconstant   = 1  a 100 = 0.002765 × 100 cellconstant = 0.2765 cm −1

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5.72 Engineering Chemistry (ii) Specific conductivity of a 0.12 N solution of an electrolyte is 0.024 ohm−1 cm−1. Determine its equivalent conductivity. Solution Equivalent conductivity (Leq ) Λ eq = K c ×

1000 1000 = Kc × Noramality Ceq

Specific conductivity (K) = 0.024 ohm−1 cm−1 1000 Normality 1000 = 0.024 × 0.12 Λ eq = 200 ohm −1 cm 2 eq −1 Λ eq = K ×

(iii) The resistance of a 0.1 N solution of an electrolyte in a conductivity cell was found to be 245 ohms, calculate the equivalent conductivity of the solution if the electrode in the cell were 2 cm apart and each has an area of 3.5 cm 2. Solution Specific conductivity K = C × =

l a

1 2 × 245 3.5

1000 Normality 1 2 1000 = × × 245 3.5 0.1 Λ eq = 23.32 ohm −1 cm 2 equ −1

Equivalent conductivity (Λ eq ) = K ×

(iv) If the equivalent conductivities at infinite dilution of NaCl, HCl and CH3COONa are 126.4, 426.1 and 91.0 ohm−1 cm 2 equiv−1 respectively, what will be the equivalent conductivity at infinite dilution for acetic acid. Solution According to Kohlrausch’s law Λ 0 for CH 3 COOH = l 0 (CH 3 COO − ) + l 0 ( H + ) Given that

Λ 0 ( NaCl) = l 0 ( Na + ) + l 0 (Cl − ) (i)

Λ 0 ( HCl) = l 0 ( H + ) + l 0 (Cl − ) (ii)

Λ 0 (CH 3 COONa) = l 0 (CH 3 COO− ) + l 0 ( Na + ) (iii)

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Λ 0 ( NaCl) = 126.4 ohm −1 cm 2 equiv −1 Λ 0 ( HCl) = 426.1 ohm −1 cm 2 equiv −1

Λ 0 (CH 3 COONa ) = 91.0 ohm −1 cm 2 equiv −1 Adding equations, (ii) and (iii) and subtracting (i), we get l0(H+) + l0(Cl−) + l0(CH3COO−) + l0(Na+) - l0(Na+) - l0(Cl−) = 426.1 + 91.0 – 126.4 l0(CH3COO−1) + l0(H+) = 390.7 ohm−1 cm2 eq−1 i.e., Λ 0 for CH 3 COOH = 390.7 ohm −1 cm 2 eq −1

(v) From the following equivalent conductivities at infinite dilution, Λ 0 for Ba (OH)2 = 457.6 ohm −1 cm 2 eq −1 Λ 0 for BaCl 2 = 240.6 ohm −1 cm 2 eq −1 Λ 0 for NH 4 Cl = 129.8 ohm −1 cm 2 eq −1

Calculate Λ 0 for NH4OH Solution

Λ 0 [Ba(OH)2 ] = l 0 ( Ba 2 + ) + 2 l 0 (OH − )

(i)

Λ 0 [Ba(Cl)2 ] = l 0 ( Ba 2 + ) + 2 l 0 (Cl − )

(ii)

Λ 0 [ NH 4 Cl] = l 0 ( NH 4+ ) + l 0 (Cl − )

(iii)

Λ 0 ( NH 4 OH) = l 0 ( NH 4+ ) + l 0 (OH − )

(iv)

1 1 eqn.(i) + eqn.(iii) − equ (ii) 2 2 1 1 = × 457.69 + 129.8 − × 240.6 2 2 Λ 0 ( NH 4 OH) = 238.3 ohm −1 cm 2 eq −1 (vi) The equivalent conductivity at infinite dilution of KCl, HCl and CH3COO K are 130.1, 379.4 and 95.6 ohm−1 cm 2 eq−1 respectively. Calculate equivalent conductivity at infinite dilution for CH3COOH. If equivalent conductivity of a given acetic acid solution is 48.5 ohm−1 cm 2 eq−1 at 25 °C. Calculate the degree of dissociation of CH3COOH at this temperature. Solution

Λ 0 for KCl = 130.1 ohm −1 cm 2 equ −1 Λ 0 for HCl = 379.4 ohm −1 cm 2 equ −1 Λ 0 for CH 3 COOK = 95.6 ohm −1 cm 2 equ −1 Λ 0 for CH 3 COOH = l 0 (CH 3 COO − ) + l 0 ( H + )

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5.74 Engineering Chemistry Λ 0 ( KCl) = l 0 ( K + ) + l 0 (Cl − )

Λ 0 ( HCl) = l 0 ( H ) + l 0 (Cl )

−

(ii) +

Λ 0 (CH 3 COOK ) = l 0 (CH 3 COO ) + l 0 ( K )

(ii) + (iii) - (i)

(i)

−

(iii)

Λ 0 (CH 3 COOK ) + Λ 0 ( HCl) − Λ 0 ( KCl) = Λ 0 (CH 3 COOH ) Λ 0 (CH 3 COOH ) = 379.4 + 95.6 − 130.1

= 344.9 ohm −1 cm 2 mol −1

Give that Λ c = 48.5 ohm −1 cm 2 mol −1 Λ Degree of dissociation (a ) = c Λ0 48.5 = 344.9 a = 0.141 (vii) At 291 K the equivalent conductivities at infinite dilution of NH4Cl, NaOH and NaCl are 129.8, 217.4 and 108.9 ohm−1 cm 2 respectively. If the equivalent conductivity of a 0.01N solution of NH4OH is 9.33 ohm−1 cm 2, what is the percentage dissociation of NH4OH at this dilution? Also calculate the dissociation constant of NH4OH. Solution Here, we are given: Λ 0 for NH 4 Cl = 129.8 ohm −1cm 2 Λ 0 for NaOH = 217.4 ohm −1cm 2 Λ 0 for NaCl = 108.9 ohm −1cm 2 By Kohlrausch’s law Λ 0 for NH 4 OH = l 0 ( NH +4 ) + l 0 (Cl − ) = Λ 0 ( NH 4 Cl) + Λ 0 ( NaOH) − Λ 0 ( NaCl) = 129.8 + 217.4 − 108.9 Λ 0 for NH 4 OH = 238.3 ohm −1cm 2 Given that, Λ 0 for NH 4 OH = 9.33 ohm −1cm 2 Λc Λ0 9.33 = 238.3 = 0.0392

∴ Degree of dissociation (a ) =

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Percentage dissociation = 0.0392 × 100 = 3.92 %

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Calculation of dissociation constant Ca 2 1−a Here, C = 0.01N a = 0.0392 K=

0.01 × (0.0392)2 1 − 0.0392 0.01 × (0.0392)2 = = 1.599 × 10 −5 0.9608

So, K =

So, dissociation constant, K = 1.599 × 10−5 (viii) The conductivity of a saturated solution of AgCl at 288 K is found to be 1.382 × 10−6 ohm−1 cm−1. Find it solubility. Given ionic conductances of Ag+ and Cl− at infinite dilution are 61.9 ohm–1 cm2 eq−1 and 76.3 ohm−1 cm2 eq−1 respectively. Solution Λ 0 ( AgCl) = l 0 ( Ag + ) + l 0 (Cl − ) = 61.9 + 76.3 = 138.2 ohm −1cm 2 eq −1 Solubility, s = K ×

1000 Λ0

= 1.382 × 10 −6 × = 10 −5 eq L−1

1000 138.2

Equivalent weight of CH3 COOH = 143.5g So, solubility = 143.5 × 10−5 g L−1 Solubility = 1.435 × 10−3 gL−1 (ix) Calculate the standard EMF of a cell which involves the following cell reaction Zn + 2Ag+→ Zn2+ + 2Ag Given that E°(Zn2+, Zn) = -0.76 volt E° (Ag+, Ag) = 0.80 volt Solution At L.H.S, Zn → Zn2+ + 2e− (oxidation) At R.H.S, 2Ag+ + 2e−→ 2Ag (Reduction) E° = E°right - E°left = 0.80 – (-0. 76) E° = 1.56 volts

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5.76 Engineering Chemistry (x) Can a solution of 1M CuSO4 be stored in a vessel made of nickel metal? Given that E°(Nl2+, Ni) = -0.25 volt E°(Cu2+,Cu) = +0.34 volt Solution In this problem, we want to see. Ni + CuSO4→ NiSO4 + Cu The cell may be represented as Ni Ni 2+ Cu 2+ Cu E° = E°right - E°left = 0.34 – (-0.25) E° = 0.59 volt Thus EMF of the cells comes out to be positive. It means CuSO4 reacts with nickel. Hence, CuSO4 cannot be stored in nickel vessel. (xi) Calculate the EMF of a Daniel cell at 25 °C, when the concentration of ZnSO4 and CuSO4 are 0.001M and 0.1M respectively. The standard potential of the cell is 1.2 volts. Solution The cell may be represented as Zn(s)  Zn 2+ (0.001M )  Cu 2+ (0.1M ) Cu (s) E cell = E°cell −

Cu 2 +  0.0591 log  2 +  n  Zn 

E°cell = 1.2 n=2 0.0591 (0.1) log 2 (0.001) 0.0591 = 1.2 + ×2 2 = 1.2 + 0.0591 = 1.2591 volt = 1.2591 volt

So, E cell = 1.2 −

E cell

(xii) Calculate the equilibrium constant for the reaction Zn + Cd 2 + Zn 2 + + Cd

 E° = 0.36 V   cell 

Solution 0.0591 log K eq n For the given reaction n = 2 E°cell =

E°cell = 0.36 V

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0.0591 log K eq 2 0.36 × 2 log K eq = = 12.1827 0.0591 K eq = Antilog(12.1827) So, 0.36 =

K eq = 1.52 × 1012 (xiii) Calculate the emf of a concentration cell at 25 °C consisting of two Zn electrodes immersed in solutions of Zn 2+ ions of 0.1 M and 0.01 M respectively. Solution EMF of concentration cell at 25 °C is given by Ecell =

C 0.0591 log 2 C1 n

[C2 > C1 ]

Now for Zn 2+/Zn, n = 2 Ecell =

0.0591 0.1 log 2 0.01

0.0591 log 10 = 0.0296 × 1 2 Ecell = 0.0296 volt

(xiv) Find the valency of mercurous ions with the help of following cell: 0.002 N mercurous 0.02 N mercurious Hg Nitrate in 0.1N HNO3 nitrate in 0.1NHNO3 Hg Anode Cathode When the emf measured at 18 °C is 0.029 volt. C 0.0591 log 2 C1 n 0.0591 0.02 0.029 = log n 0.002 Ecell =

By solving, n = 2.

(xv) While determining the pH of a solution, the quinhydrone electrode, H+, Q, QH2 was used in conjunction with a saturated calomel electrode as represented below; Hg,Hg 2 Cl 2 (s); KCl(Sat. Soln)  Η + ( unknown); Q, QH 2 , Pt The EMF of the cell was found to be 0.2640 volt at 25 °C. Calculate the pH of the solution at this temperature given that Ecalomel = +0.2422 volt at 25 °C and E°(H+, Q, QH2) = +0.6996 volt

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5.78 Engineering Chemistry Solution The EMF of the cell is given by E = Eright - Eleft 0.2640 = 0.6996 – 0.0591 pH – 0.2422 pH =

0.6996 − 0.2422 − 0.2640 0.0591

pH = 3.27

5.16.5 Descriptive Questions Q.1 On the basis of electrolytic condition, how are the electrolytes classified. Q.2 What is an electrochemical series? Discuss its important applications. Q.3 Explain Kohlrausch’s law of independent migration of ions. Mention one application of kohlrausch’s law. Q.4 What are concentration cells? Explain with example? Q.5 What is an ion selective electrode? Explain its principle and working. Q.6 Explain the following terms. a. Specific conductance b. Molar conductance c. Equivalent conductance. What will be effect of dilution on them? Q.7 Discuss briefly conductometric titration. Q.8 a. In conductrometric titration, more concentrated solution is added from burette, why? What are the advantages of conductrometric titration over ordinary volumetric methods? b. Draw weak acid – strong base conductometric titration curve and explain it. Q.9 What is standard electrode potential? Give its importance. Q.10 a. Why do electrochemical cells stops working after some time? b. W hy does blue color of copper sulphate solution fade when it is electrolyzed using platinum electrodes? Q.11 What are redox electrodes? Indicate the electrode reaction and electrode potential with suitable example. Q.12 What is electrochemical series, Give its applications with suitable examples. Q.13 Describe the following electrodes giving the diagram, electrode notation and electrode reaction (i) Standard hydrogen electrode (ii) Calomel electrode. Q.14 Write short note on fuel cell, how it is different from commercial cell? Mention the advantages of fuel cells.

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Q.15 Write briefly about (i) Primary cells (ii) Secondary cells. Q.16 Explain the working principle of methanol - oxygen fuel cell. Q.17 Explain the working principle of Ag - AgCl electrode. Q.18 Describe the construction of lead-acid battery with the reaction occurring during discharge. Q.19 How does fuel cell differ from battery? Q.20 What are reference electrodes? Explain the working of quinhydrone electrode. Q.21 Explain the composition, applications and advantages of the following cells (i) Ni-Cd Cell (ii) Lithium ion cell (iii) Dry cell Q.22 How are specific conductance and equivalent conductance related to concentration of an electrolyte? Q.23 Explain the measurement of pH of solution using glass electrode. Mention the advantages of this electrode. Q.24 What is an ion-selective electrode? Explain its working. Q.25 Explain the construction and functioning of a Daniel cell. Q.26 Explain the EMF method for determination pH of a solution. Q.27 What is emf? How is it measured potentiometrically? Q.28 Derive Nernst equation for the calculation of cell emf. Q.29 Differentiate primary, secondary and fuel cells with examples. Q.30 How do you differentiate between EMF series from galvanic series?

5.16.6 Problems for Practice 1. The resistance of 0.01 N NaCl solution at 25 °C is 200 ohms. Cells constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution. [Ans.: 500 ohms−1 cm2] 2. Molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2, what would be the specific conductance of this solution. [Ans.: 0.208 S cm−1] 3. The measured resistance of a conductance cell containing 7.5 × 10−3 M solution of KCl at 25 °C was 1005 ohms. Calculated (a) specific conductance and (b) molar conductance of this solution. Given that cell constant = 1.25 cm−1 [Ans.: (a) = 0.001244 S cm−1 (b) = 165.87 S cm 2 mol−1]

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5.80 Engineering Chemistry 4. The resistance of a decinormal solution of a salt occupying a volume between two platinum electrodes 1.8 cm apart and 5.4 cm 2 in area was found to be 32 ohms. Calculate equivalent conductance of this solution [Ans.: 104.1 ohm−1 cm2 eq−1] 5. The resistance of 0.2N solution of an electrolyte was found to be 250 ohms at 25 °C. Calculate the equivalent conductivity if the cell constant is 0.75 cm−1 [Ans.: 15 ohm−1 cm2 eq−1] 6. Calculate the equivalent conductance at infinite dilution for CH3COOH, given that Λ 0 ( HCl) = 425 ohm −1 cm 2 eq −1 Λ 0 ( NaCl) = 188 ohm −1 cm 2 eq −1 Λ 0 (CH 3 COONa ) = 96 ohm −1 cm 2 eq −1 [Ans.: 333 ohm−1 cm2 eq−1] 7. The equivalent conductance of NaOH, NaCl and BaCl2 at infinity dilution are 2.481 × 10−2, 1.265 × 10−2 and 2.800 ×10−2 ohm−1 m 2 eq−1 respectively. Calculate Λ 0 for Ba (OH)2. [Ans.: 5.232 × 10−2 ohm−1 m 2 eq−1] 8. If the equivalent conductivities at infinity dilution at 293 K for HCl, CH3COO Na and NaCl are 383.5, 78.4 and 102.0 ohm−1 cm 2 respectively. Calculate equivalent conductivity at infinity dilution. The equivalent conductivity of CH3COOH at other dilution is 100.0 ohm−1 cm 2 at 293 K, Calculate degree of ionization of acetic acid at this dilution. [Ans. : Λ 0 = 359.9 ohm −1cm 2 ; a = 0.278] 9. The specific conductivities of a saturated solution of AgCl is 2.30 × 10−6 ohm−1 cm−1 at 25 °C. Calculate the solubility of AgCl at 25 °C, Given that l 0 ( Ag + ) and l0(Cl−) are 61.9 and 76.3ohm−1cm 2 eq−1 respectively. [Ans.: 2.388 × 10−3 gL−1] 10. Calculate the percentage dissociation of AgNO3 at 18 °C, given that ionic conductivity of Ag+ and NO3− ions are 56.7 and 60.5 ohm−1 cm 2 eq−1. The specific conductance of a decinormal solution of Ag NO3 at 18 °C is 0.0085 ohm−1 cm−1 [Ans.: 72.52%] 11. The equivalent conductivity of 0.025 N HCOOH acid is 46.1 S cm 2 eq−1. Calculation its degree of dissociation and dissociation constant. Give that l0 ( H + ) = 349.65 cm 2 eq −1

l0 ( HCOO − ) = 54.65 cm 2 eq −1

[Ans.: degree of dissociation, 0.114, Dissociation const, k = 3.67 × 10−4] 12. For the cell show below: Zn(s) ZnSO 4 (aq) CuSO 4 (aq ) Cu (s)

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Calculate standard cell potential if standard state reduction electrode potential for Cu2+/Cu and Zn 2+/Zn is +0.34 V and – 0.76 V respectively. [Ans.: 1.10 V] 13. Can we use a copper vessel to store 1 M Ag NO3 solution? Given that E °Cu 2+ | Cu = +0.34 V E° Ag+ | Ag = +0.80 V [Ans.: No, we can’t use copper vessel to store Ag NO3] 14. A galvanic cell consist of a metallic zinc plate immensed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Given that standard emf of the cell is 0.63 volts. Also represents the cell reaction. [Ans.: EMF = 0.6094 volts, cell reaction Zn | Zn 2+ || Pb2+| Pb] 15. Calculate the EMF of the following cell at 298 K Cd(s) | cd 2+ (0.04M) || Ni2+(2.0M) | Ni (s) given that standard EMF of the cell is 0.15 volt. [Ans.: Ecell = 0.20 volt] 16. Calculate the EMF of the following cell Cr | Cr3+ (0.1M) || Fe2+(0.01M) | Fe Given that standard EMF of the cell is 0.30 volts [Ans.: Ecell = 0.2606 volts] 17. Calculate the equilibrium constant for the reaction Sn 2+ (aq ) + Pb(s) Sn(s) + Pb 2+ (aq) at 298K Given that E° Sn2+/Sn = -0.14 V E° Pb2+/Pb = -0.13V [Ans.: Keq = 0.46] 18. Calculate the standard emf of the H2–O2 fuel cell, given E° values as -0.40 V and 0.83 V for hydrogen and oxygen half-cell respectively. [Ans.: 1.23 V] 19. Calculate the EMF of the electrode-concentration cell Pt; H2(p1), HCl, H2(p2); Pt at 25 °C If p1 = 600 torr and p2 = 400 torr, [Ans.: Ecell = 5.19 × 10−3 V] 20. The EMF of the following cell at 25 °C is 0.445 volt. The cell reaction is represented as pt,H 2 (g) H + ( unknown conc) KCl (sat. soln.) Hg 2 Cl 2 , Hg Calculate the pH of the unknown solution, give that reduction potential of saturated calomel electrode is 0.2422 volt. [Ans.: pH = 3.38]

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Science of Corrosion

6.1 Introduction Corrosion can be defined as the slow degradation or deterioration of a metallic material from the metallic surface due to unwanted attack by the atmosphere gases, soil, chemical, or electrochemical reaction with its environment (gaseous or liquid medium). Degradation or deterioration means reduction in the useful properties of the material which includes: (i) Decaying of surfaces of metals (ii) Weakening of the material due to loss of cross sectional area. (iii) Loss of properties such as malleability, ductility. (iv) Cracking of polymer due to sunlight.

6.1.1 Causes of Corrosion It has been found that most metals (exceptions noble metals like Au, Pt, etc.) exist in nature is combined forms as their oxides, carbonates, sulphates, sulphides, chlorides etc. Metals are extracted from their ores by using different extraction processes. Energy is required for the extraction of metals. So, consequently pure metals have higher energy than combined form which has lower energy. For this, metals easily undergo interaction with their environment either chemically or electrochemically to form a Stable compound by the process of corrosion. Corrosion is an oxidation process in which metallic compound having lower energy is formed and energy liberates. Metal

Corrosion (Oxidation) Metallic compound + Energy Metallurgy(Reduction)

( Higher energy)

(L Lower energy )

 Thermodynamically    Unstable  

 Thermodynamically    Stable  

Examples: (i) Rusting of iron: When Iron exposed to the atmospheric conditions, a layer of reddish scale and powder of Fe3O4 is formed on the surface.

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6.2 Engineering Chemistry (ii) Formation of green film on the surface of copper: A green layer of basic carbonate consisting of [CuCO3 + Cu (OH)2] is formed on the surface of copper when exposed to moist air. (iii) Tarnishing of silver: When silver is exposed to the atmosphere, a black coating of air is formed.

6.1.2 Types of Corrosion Various types of corrosion processes along with their respective mechanism are given below:

Corrosion

Dry or chemical

Wet or Electrochemical

Atmospheric

Corrosion Other Liquid by gases Metal Oxygen

Galvanic

Pitting

Stress

Soil

Other forms

Inter Granular Erosion Microbiological

Crevice Concentration Water line Cell or Differential aeration

Dry or Chemical Corrosion In such a type, corrosion occurs due to direct chemical action of atmospheric gases such as oxygen, halogens, sulphur dioxide and hydrogen sulphide with metals resulting into the formation of compounds such as oxides, halides, sulphates and sulphides is known as chemical corrosion. The products which are formed are insoluble, soluble or liquid in nature. Dry Corrosion is of types (i) Oxidation corrosion (Corrosion by oxygen) (ii) Corrosion by other gases (iii) Liquid metal corrosion (i) Oxidation corrosion: Oxidation corrosion is due to the direct chemical attack of oxygen on the metal in the absence of moisture at low or high temperature leading to the oxidation of metal. Alkali metals (Li, Na, K, etc) and alkaline earth metals (Be, Ca, Sr, etc) are oxidized at low temperatures, whereas all other metals (except Ag, Au, and Pt) are oxidized at high temperatures.

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6.3

Mechanism When a metal is exposed to air, absorption of oxygen takes place even at ordinary temperatures. This absorption is purely physical in nature and is due to vander Waal’s forces. However, due to climatic changes, the absorbed oxygen may gradually enter into chemical combination with the metal by electron transfer between the metal atoms and oxygen as shown below: 2M → 2M n + + 2 ne − (metal ion )

(loss of electrons by metal)

n O + 2 ne − → nO2 − (gain of electrons by oxygen) 2 2 (oxide ion ) overall reaction

2M +

n O → 2M n+ + nO2 − → M 2 O n 2 2 ( metal ion) oxide ion

metal oxide

The metal oxide scale is formed at the metal surface. This scale acts as a barrier and tends to prevent the underlying metal atoms to come in contact with oxygen. The continuation of the oxidation process depends upon two factors.

(a) The nature of the oxide film formed (b) The rate of diffusion of the metal ion and oxide ion through the layer formed.

Nature of the Oxide Film Formed on the Surface hen a metal is placed in atmosphere a thin layer of oxide film is formed at the surface of the metal W which can be written as Metal + Oxygen → Metal oxide This metal oxide layer can be (i) Stable: When the oxide film is stable, impervious and highly adhering, such kind of layer forms a shield for metal surface. The layer consists of fine grain particles which tightly sticks to the metal surface and does not allow oxygen to diffuse into the metal surface and thus prevents metal from corrosion e.g. Al, Pb, Cu, Sn etc. (ii) Unstable: When the oxide film is unstable and has tendency to decompose back to metal and oxygen, it does not undergo in oxidation corrosion e.g. Au, Ag, Pt, etc. Metal oxide Metal + Oxygen (iii) Porous: When the oxide layer having pores or cracks. In such a case, diffusion of cations (Mn+) and anions (O2-) take place smoothly then oxidation corrosion takes place continuously, till the entire metal is completely converted into its oxide. Porous Metal Oxide Exposed surface Metal

+O2 (of air)

Metal

Further attack on metal surface through pores/cracks

The porous nature of oxide film may be explained by pilling-Bedworth rule.

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6.4 Engineering Chemistry Pilling Bedworth Rule This rule describes the protective and non-protective nature of the oxide layer which is formed during corrosion. According to this rule, the specific volume ratio is calculated as follow: Volume of metal oxide Specific volume Ratio = Volume of metal (i) If the specific volume ratio is smaller, the oxidation corrosion will take place because the oxide films will be sufficiently porous for diffusion of M n+ and O2-. (ii) If volume of metal oxide ≥ volume of parent metal, then it will be non-porous or protective. (iii) If volume of metal oxide < volume of parent metal, then it will be porous or non-protective. Examples: Alkali and alkaline earth metals (like Li, Na, K, Mg) form oxides of volume less than the volume of metal. So, oxide layer faces stress and strains, which result in development of cracks and pores in its structure. So, further corrosion continues till the whole metal is destroyed. But in case of metal like Al forms oxide, whose volume is greater than volume of metal. So, nonporous, tightly adhering layer is formed, so rate of oxidation rapidly decreases to zero. Rate of Diffusion of Metal Ion and Oxide Ion Through the Layer Formed Metal and oxygen combine to form metal oxide which forms a thin film whose thickness is less than 300A°, and it’s called as scale, if its thickness exceeds this value. This film or scale prevents further oxidation. But for oxidation to continue either the metal ion must diffuse outwards through the scale to the surface or oxygen ion must diffuse inwards through the scale to the underlying metal. Both transfers occur, but outward diffusion of metal ion is much easier because metal ions are smaller than oxide ion and of higher mobility as shown in Figure 6.1. Reaction at metal-Metal oxide interface M → Mn++ ne−(oxidation)

Inward diffusion of oxide ion through scale (slow)

O2− Mn+

Atmospheric oxygen (air)

O2−

Formation of metal oxide (M2O) at the point of meeting of ions

Mn+

Metal (M)

e−

Reaction at exposed part 1 O + 2e−→O2− (Reduction) 2 2

Direction of electron flow

Outward diffusion of metal ion through scale (fast process)

Figure 6.1 Oxidation mechanism of metals

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6.5

(ii) Corrosion by other gases: Corrosion also occurs by other gases like SO2, Cl2, CO2, H2S, F2 etc. This depends upon the affinity of metals with these metals. The degree of attack depends on the formation of protective (non-porous) or non-protective (porous) film on the surface. The extent of corrosion depends upon the following: (a) Nature of the environment: The environment plays very important role in corrosion because it facilitates the affinity between metal and gases. (b) Chemical affinity between metal and gas: If the affinity between metal and gas is more, then corrosion will be more and more (c) Nature of the film formed on the metal surface: (1) If the film formed is protective or non-porous,then the intensity of attack decreases Eg: 2Ag + Cl 2 →

2Ag Cl ( protective layer)

(2) If the formed is non–protective or porous, metal is destroyed rapidly Eg: Sn + 2Cl 2 →

SnCl 4

Volatile layer i.e non-protective layer

It evaporates and metal surface is exposed for corrosion. (iii) Liquid metal corrosion: This type of corrosion happens when liquid metal flows over solid metal or alloy at high temperature and solid metal or alloy usually gets weakened. This type of corrosion mainly occurs in nuclear powers devices. There are two possibilities of liquid metal corrosion: (a) Either the liquid dissolves the solid metal surface. (b) Liquid penetrates into the solid surface and thus weakens the bond. Wet or Electrochemical Corrosion It is also known as immersed corrosion. It is more common than dry corrosion. It occurs mostly under wet or moist conditions through the formation of electrochemical cells, and is therefore, referred to as electrochemical corrosion. Wet corrosion can be easily explained by electrochemical theory. Electrochemical Theory of Corrosion All metals have tendency to pass into solution. The tendency of metal to pass into solution when immersed in a solution of its salt is measured in terms of electrode potential. If a metal having lower reduction potential (higher electropositive) comes into contact with another metal having a higher electrode potential (higher electro negative) a galvanic cell is set up. The metal having lower electrode potential becomes anodic and get dissolved as corresponding metallic ions with the liberation of free electrons. M → Mn+ + ne- (Oxidation) The metal with high electrode potential acts as cathode and gets protected during the process.

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6.6 Engineering Chemistry Mechanism of Electrochemical Corrosion (i) The existence of separate cathodic and anodic areas between which current flows through the conducting solution. (ii) Oxidation (loss of electrons) takes place at the anodic area and the metal is destroyed by either dissolution or combination with oxygen. Hence, corrosion always takes place at the anode. M → Mn+ + ne- (Oxidation) (iii) Reduction (gain of electrons) takes place at the cathode. The electrons from the anode are a ccepted by the dissolved oxygen forming ions such as OH- or O2- ions. 1 O + H 2 O + 2e − → 2OH − 2 2

( Reduction )

or O2 + 2e − → 2O2 −

( Reduction )

(iv) The metallic ions (at anodic area) and non-metallic ions (at cathodic area) diffuse towards each other through conducting medium and form a corrosion product somewhere between anode and cathode. Depending on the nature of corrosive environment, the mechanism of electrochemical corrosion may be explained in terms of (i) Evolving of hydrogen (ii) Absorption of oxygen (i) Evolution of hydrogen: The process of corrosion in which H2 is liberated is called evolution of hydrogen type corrosion. This mechanism of corrosion follows usually in acidic environment. Thus, in acidic medium (absence of oxygen) hydrogen ion acquire electrons with the liberation of H 2 gas in cathodic reaction and the anode is the metal which undergo oxidation and looses electrons to the environment and pass into solution. This process is shown in Figure 6.2. For example: (a) If iron metal is used, the dissolution of iron as Fe2+ Fe → Fe2+ + 2e- (Oxidation) (b) These electrons flow through the metal from anode to cathode, Where H+ ions of acidic solution accept these electrons and get reduced in the form of H2 gas. 2H+ + 2e- → H2 (Reduction) Overall Reaction Fe + 2H+ → Fe2+ + H2

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Science of Corrosion

H+

2H++ 2e− → H2 Fe → Fe2++2e−

Acidic Solution (Electrolyte)

H+

Cathodic reaction

6.7

H+ Fe → Fe2++2e−

Anodic Reaction Flow of electrons

Anodic area (large)

Small Cathodic area

Anodic area (large)

Iron Metal

Figure 6.2 Mechanism of wet corrosion by hydrogen evolution It is important to note that in hydrogen evolution type of corrosion, anodic areas are very large in comparison to cathodic areas. All the metals, above hydrogen in electrochemical series have a tendency to get dissolved in acidic solution with liberation of hydrogen. (ii) Absorption of oxygen: This type of corrosion occurs in basic or neutral environment (such as NaCl solution used as electrolyte). The common example is corrosion of iron occurs by oxygen in the presence of aqueous solution of NaCl in the presence of oxygen. This process is shown in Figure 6.3.

Rust Fe → Fe2++2e− 1 O 1 O + H O + 2e− − + H O + 2e 2 2 2 2 2 2 − ↓ 2OH ↓ 2OH−

Aqueous neutral solution of electrolyte (NaCl)

Iron oxide film Cathodic Area (large)

Anodic Area (small)

Cathodic Area (large)

Flow of electrons

Figure 6.3 Mechanism of wet corrosion of absorption of oxygen (a) At anode, iron dissolves to form ions as Fe → Fe2+ + 2e- (Oxidation) (b) At cathode, the electrons evolved by above reaction are accepted by oxygen in presence of water. 1 O + H 2 O + 2e − → 2OH − ( Reduction ) 2 2

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6.8 Engineering Chemistry (c) The Fe2+ ions (at anode) and OH- (at cathode) diffuse and when they combine Fe(OH)2 is precipitate. Fe2+ + 2OH- → Fe(OH)2↓ (d) In the presence of sufficient oxygen, Fe(OH)2 can be easily oxidized into ferric hydroxide [Fe(OH)3] 4 Fe (OH)2 + 2H 2 O + O2 → 4Fe(OH)3 ↓ Yellow rust

(e) If the supply of oxygen is limited then black anhydrous magnetite i.e. ferrousoferic oxide is formed as 3Fe(OH)3 → Fe3O4·6H2O or Fe2O3·FeO·6H2O(Black rust) Difference between chemical and electrochemical corrosion Chemical corrosion

Electrochemical corrosion

1. It occurs only in dry conditions.

It occurs in wet conditions in the presence of moisture and electrolyte. It involves electrochemical attack of corrosive environment on the surface of metal. Corrosion products accumulate some where between the area of anode and cathode. It is a continuous process. In this process, oxidation and reduction takes place at different sites. It is a fast electrochemical process. It proceeds through the cells. In electrochemical corrosion product is always stable e.g. Fe3O4, Zn(OH)2 etc. In this process, path for electron flow is always required. Its mechanism is explained on the basis of electrochemical reaction.

2. It involves chemical attack of oxygen or other gases 3. Chemical corrosion products accumulate at the site of attack i.e. at anode 4. It is a self-controlled process. 5. In this process, oxidation and reduction sites are same. 6. Chemical corrosion is a slow process taking place by chemical reaction of atmospheric gases. 7. In chemical corrosion, product may be unstable, volatile or porous in nature. 8. In chemical corrosion, path for electron flow is not required. 9. It is explained on the basis of absorption.

Types of Electrochemical Corrosion Such type of corrosion takes place in following conditions: (i) When two dissimilar metals or alloys are in contact with each other in the presence of a conducting medium (aqueous solution, moisture etc.) (ii) Separate anodic and cathodic areas between which the current flows the conducting medium. (iii) Oxidation takes place at anode and reduction takes place at cathode e.g. rusting of iron. (a) Galvanic corrosion or bimetallic corrosion The galvanic cell is formed if two different metals (e.g. zinc and copper) are electrically connected and exposed to an electrolyte. As a result, the less noble metal (i.e. the metal having a lower value of standard reduction potential or placed higher in the electrochemical series) gets corroded. This type of corrosion is called as galvanic corrosion e.g.: Zn-Cu, Zn-Ag, Fe-Cu etc.

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6.9

In Zn-Cu galvanic cell, Zinc (E° = −0.76 V) with lower reduction potential than copper (E° = + 0.34 V) acts as anode and the electrons flow from anodic metal (Zn) to cathodic metal (Cu). The anodic metal is corroded, while cathodic metal remains protected. This process is shown in Figure 6.4. Zn (anode) (less noble)

Cu(cathode) (More Noble)

e−

Conducting Solution

−

Zn → Zn2++2e

Figure 6.4 Galvanic corrosion (The less noble metal zinc acts as anode and undergoes corrosion, whereas the most noble metal copper remains protected.)

(1) In acidic solution, the corrosion occurs by evolution of hydrogen Zn → Zn 2+ + 2e −

At anode: At cathode :

−

2 H + 2e → H 2 +

(Oxidation ) ( Reduction )

Zn + 2H + → Zn 2+ + H 2

(2) In Neutral or slightly alkaline medium, the corrosion occurs by absorption of oxygen At anode :

Zn → Zn 2 + + 2e − (Oxidation )

1 H 2 O + O2 + 2e − → 2OH − 2 At cathode : 1 Zn + H 2 O + O2 → Zn(OH)2 2

( Reduction )

Examples: (i) Steel screw’s in a brass marine hardware (ii) A steel propeller shaft in bronz bearing. (iii) Steel pipe connected to copper plumbing.

Control Since Galvanic depends upon the following factors: (i) Greater the potential difference between two metals, greater is the corrosion. (ii) Suitable medium for corrosion (iii) Surface area of the metal Hence corrosion may be controlled by the following factors: (i) Avoiding the suitable medium for corrosion. (ii) Minimizing the potential difference of metals i.e. avoiding the galvanic couples. (iii) By polishing the metals.

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6.10 Engineering Chemistry Concentration Cell Corrosion (Differential Aeration Corrosion) When a metal is exposed to an electrolyte of varying concentrations or to varying aeration, it undergoes an electrochemical attack due to formation of miniature concentration cells on its surface and gets corroded. Differential aeration corrosion is the most common type of concentration cell corrosion. This type of cell is formed when the metal is kept in different air concentration i.e. two ends of metal surface are at different concentration of air. The part of metal which is poorly oxygenated acts as anode and other part of the metal which is highly oxygenated acts as cathode. This develops electrode potential and hence the, metal gets corroded. Let us consider the case of zinc (Zn) rod which is immersed in NaCl solution. A potential difference is developed between differently aerated areas. The part of the rod which is at greater depth acts as anode (less oxygenated) and that is which is above the surface acts as cathode (more oxygenated) and zinc corrodes due to electrode potential. This process is shown in Figure 6.5.

Zn rod Cathode

Anode Flow of electrons

e− Corroding anode

Zn2+

Zn → Zn2++2e

Electrolyte NaCl

Zn2++ 2Cl− → ZnCl2

Figure 6.5 Differential aeration corrosion At anode: less oxygenated part Zn → Zn 2+ + 2e- (Oxidation) At cathode; more oxygenated part 1 O + H 2 O + 2e − → 2OH − 2 2

( Reduction )

Overall Reaction Zn 2+ + 2OH- → Zn (OH)2

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In a similar way, iron metal corrodes under drop of water (or salt solution). Areas covered by droplets, having no access of oxygen, it become anodic with respect to the other areas, which are freely exposed to air become cathodic. This process is shown in Figure 6.6. Air Good access of oxygen

Fe → Fe2++2e−

Drop of salt solution Rust ring (Iron hydroxide) Cathodic area (Protected)

Cathodic area (Protected) Anode (Poor access of oxygen attacked) Iron

Figure 6.6 Differential aeration corrosion At anode: Fe → Fe2+ + 2e- (Oxidation) 1 − − At cathode: O2 + H 2 O + 2e → 2OH (Reduction) 2 1 Overall reaction Fe + O2 + H 2 O → Fe(OH) 2 2 Iron hydroxide

Important Characteristic about Differential Aeration Corrosion (i) The metal having low oxygen concentration part act as anode and metal having high oxygen concentration act as cathode. (ii) Corrosion may be accelerated in apparently in accessible places, because of deficiency of oxygen at some part. (iii) This type of corrosion also accelerated under accumulation of dirt, sand, scale or other contamination, because such covered part act as anode due to difference in air concentration. (iv) It is a localized attack on some oxygen deficient areas such as metal exposed to aqueous media corrode under blocks of wood or pieces of glass, which screen that portion of metal from oxygen access, resulting into localized pitting. Water-line Corrosion It is the type of differential aeration corrosion, which occurs when a metal is partly immessed in water. The corrosion takes place just below the waterline and hence it is known as waterline corrosion. It is an observed fact that when water is kept stagnant in a steel tank for a long time, corrosion takes place just below the water level, it is due to the concentration of dissolved oxygen at the water surface is greater than that under surface.It forms an oxygen concentration cell. The area above the waterline (highly oxygenated) acts as cathodic and corrosion takes place along a line just beneath the level of water meniscus (anodic area) as shown in Figure 6.7.

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6.12 Engineering Chemistry Cathodic Part H2O + 1 O2 + 2e− → 2OH

−

Anodic part undergoing corrosion Fe → Fe2++ 2e−

Anodic Part undergoing corrosion

Water Steel tank

Figure 6.7 Water-line corrosion Corrosion takes place at anodic part At anode:

Fe → Fe2+ + 2e- (Oxidation)

At cathode:

1 H 2 O + O2 + 2e − → 2OH − (Reduction) 2

Overall reaction:

Fe2 + + 2OH − → Fe (OH)2 ↓ oxidation Fe(OH)3

Corrosion product

This type of corrosion is accelerated when water is acidic in nature and presence of salts like chlorides, bromides, etc. When marine plants attach themselves to side of the ships, this type of corrosion is increased because of presence of different salts. Prevention (i) Water-line corrosion is reduced when the water is free from acidic impurities. (ii) Usage of special anti foaming paints minimizes such type of corrosion to some extent. (iii) By using anodic inhibitors like phosphates, carbonates, silicates, water-line corrosion can be retarded. This type of corrosion is accelerated when water is acidic in nature and presence of salts like chlorides, bromides, etc. When marine plants attach themselves to side of the ships, this type of corrosion is increased because of presence of different salts. The use of special antifouling paints minimizes such type of corrosion to some extent. Pitting Corrosion Pitting corrosion is a non-uniform corrosion which is caused by localized accelerated attack on metal surface and forms pits, cavities and pin holes in the metal. A pit is formed when the protective coating on the metal surface breaks at specific points. Once the pit is formed the process of corrosion becomes fast due to differential amount of oxygen in contact with metal surface. The portion with higher concentration of oxygen become cathode and that with lower oxygen concentration becomes anode. This causes corrosion of metal. This process is shown in Figure 6.8.

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Corrosion product Cathode − H2O + 1 O2 + 2e− → 2OH

Cathode − H2O + 1 O2 + 2e− → 2OH 2

e−

Flow of electrons

Pit (Anode) Iron Fe → Fe2+ + 2e−

Figure 6.8 Pitting corrosion At anode:

Fe → Fe2+ + 2e- (Oxidation)

At cathode:

1 H 2 O + O2 + 2e − → 2OH − (Reduction) 2

Overall Reaction

1 O] O → Fe(OH)2 [ → Fe(OH)3 Rust 2 2 [O] 2+ − Fe + 2OH → Fe(OH)2 → Fe(OH)3

Fe + H 2 O +

Pitting corrosion may be caused by (i) (ii) (iii) (iv) (v) (vi)

Surface roughness Scratches on metal surface Local strains of metal due to non-uniform stress Presence of extrageneous impurities (like sand, dust or scale) Presence of drop of salt solution Non-uniform polishing of metal etc.

Pitting corrosion may be prevented by (i) Proper designing of metal (ii) Proper polishing of metal (iii) Use of pure metal Stress Corrosion Stress corrosion or stress cracking is the type of corrosion which occurs due to combined effect of tensile stresses and the corrosive environment on metal when metal is exposed to corrosive environment. Pure metal generally does not undergo stress corrosion whereas fabricated metal components or an article of certain alloys like zinc and nickel brasses undergoes such types of corrosion.

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6.14 Engineering Chemistry Favorable Conditions for Stress Corrosion (i) Tensile stress: The stress developed on metal surface may be internal or external. Internal stress is caused by manufacturing process (quenching, bending, annealing, etc.) or fabrication process or heat treatment process. In all such cases, metal under stress becomes more anodic and that area undergoes corrosion. (ii) Corrosive environment: The specific and selective environment play very important role in stress corrosion. For example (a) Mild steel undergoes stress corrosion in the presence of caustic alkalies and strong nitrate solution. (b) Stainless steel in the presence of acid chloride solution. (c) Brass in the presence of traces of ammonia. Mechanism Stress corrosion is localized electrochemical phenomenon. As we know that, the point or area under stress as well as grain boundaries act as electrochemical cell which occurs generally due to internal stresses due to metallurgical operations such as bending, pressing, rolling, quenching, annealing, etc. Due to presence of stress forms anodic areas in localized zones with respect to more cathodic areas at the metal surface. Such areas under stress act as anode and they become so chemically active that they are attacked, even by a mild corrosive environment, which result in the formation of cracks which propagate rapidly resulting in an unexpected failure of the metal surface. This process is shown in Figure 6.9. Unstressed part acting as cathode

Grain Boundary (Part under stress acting as anode)

Grain A

Grain B

Crack due to stress corrosion

Figure 6.9 Stress corrosion In every type of corrosion there is formation of galvanic cells and corrosion takes place at the anodic part. At anode:

M → Mn+ + ne –

(Oxidation)

At cathode: H 2 O + 1 O2 + 2e − → 2OH − (Reduction) 2 Stress corrosion takes place even in mild corrosive environment on the stressed metal part. Types of Stress Corrosion (i) Season cracking: This type of cracking is generally refers to the corrosion of copper alloys, particularly brass. Brasses are binary alloys of Cu and Zn which are electrochemically reactive

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in an environment of ammonia. Therefore, when brass is exposed in ammonical medium, both copper and zinc form complexes by losing electrons in ammonical solution. As a result, dissolution of brass occurs and forms cracks for stress corrosion. Zn → Zn 2+ + 2e- and Cu → Cu 2+ + 2eZn 2 + + 4 NH 3 → [Zn( NH 3 ) 4 ]2 +

and

Cu 2 + + 4 NH 3 → [Cu ( NH 3 ) 4 ]2 +

This reaction is generally referred to a season cracking. (ii) Caustic Embrittlement: This type of corrosion generally occurs in mild steel, which undergoes stress corrosion in caustic alkalies at high temperature and pressure. It is very dangerous form of stress corrosion, generally occurs in steam-boilers and heat-transfer equipments in which water of high alkalinity attack the mild steel plants, particularly crevices near rivets, bends, joints etc. The causes and methods of prevention of caustic embrittlement: For water-softening purpose of Boiler-water, we generally added a certain proportion of sodium carbonate into it. In high pressure boilers, this breaks up to give sodium hydroxide and carbon dioxide. Na2CO3 + H2O → 2NaOH + CO2↑ This makes boiler-water alkaline in nature. This dilute alkaline boiler-water flows into the minute cracks and crevices by capillary action, where water evaporates and caustic soda concentration builds up. The area where metal is stressed and concentration of alkali is much higher than that in the body of the boiler, alkali dissolve metal as sodium ferrate in crevices, cracks etc. sodium ferrate is decomposes according to either of the following reactions: 3Na2FeO2 + 4H2O → 6NaOH + Fe3O4 + H2 or 6Na2FeO2 + 6H2O + O2 → 12NaOH + 2Fe3O4 Sodium hydroxide (NaOH) is regenerated and magnetite (Fe3O4) is precipitated, thereby enhancing further dissolution of iron. Caustic embrittlement can be explained by considering the following electrochemical cell: Fe

NaOH (Concentrated ) Anode

NaOH ( Dilute)

Cathode

The iron surrounded by dilute NaOH is the cathodic area; while iron surrounded by concentrated NaOH (e.g. crevices, hair-cracks, rivets etc.) is the anodic area and undergoing corrosion and is thus dissolved the iron metal from that areas. Prevention of Caustic Embrittlement (i) Use of sodium sulphate in boiler-water. (ii) Use of tannin or lignin as additive boiler-water. Both these methods prevent caustic cracking by blocking up the cracks and crevices with innocuous harmless substances, thereby preventing the sodium hydroxide from infiltrating into these areas. (iii) Corrosion fatigue: This type as corrosion cracking occurs due to repeated stresses caused by shaking, tapping, vibration etc. in the presence of corrosive environment. The repeated stress make same

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6.16 Engineering Chemistry metal less elastic which on turn act as anode with respect to other part of metal. The corrosion take place in these region and cracks occurs. This type of corrosion occurs mostly in alloy steel.

6.2 GALVANIC SERIES Electrochemical series is very helpful to understand the extent of corrosion on the basis of standard reduction potential. According to this series a metal placed at top in the series is more anodic and undergoes corrosion rapidly than the metal below in the series. The rate and severity of corrosion depends upon the difference in their positions, greater is the difference, and the faster is the corrosion of metal. For example, Li corrodes faster than Mg; Zn corrodes faster than Fe, and so on. However, some exceptions to this generalisation have been noticed. For example, position of titanium (Ti) is higher than silver (Ag) but Ti is less reactive towards corrosion. Similarly, aluminium (Al) is above zinc (Zn) but zinc corrodes faster. This is only due to formation of strongly adhering oxide layers on their surfaces, thereby making their effective electrode potential more positive (or less negative). Hence a new series came into exist which is based on relative oxidation potential in sea water. This series is known as galvanic series. Galvanic series is shown in Table 6.1. Table 6.1 Galvanic series Active (or Anodic)

1. Mg 2. Mg alloys 3. Zn 4. Al 5. Cd 6. Al Alloys 7. Mild steel 8. Cast iron 9. Stainless steel 10. Pb-Sn alloy (solder) 11. Pb 12. Sn 13. Brass 14. Monel (Ni = 7%, Cu = 30%, Rest Fe) 15. Silver solder 16. Cu 17. Ni 18. Bronze 19. Cu-Ni alloys 20. Ag 21. Chromium stainless steel 22. Graphite 23. Ti 24. Au 25. Pt

Noble (or Cathodic)

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According to this series the metal or alloys higher up the position in the series is more anodic and undergoes corrosion very rapidly. For example, the position of Zn is higher than Al; hence Zn undergoes corrosion rapidly not aluminium Al.

6.2.1 Factors Affecting Corrosion The rate and extent of corrosion of a metal depends upon the following factors: (i) Nature of the metal and (ii) Nature of the environment. (i) Nature of the metal: The various factors such as its purity, position in galvanic series, physical state, overvoltage etc. which decide the nature of a metal from the view point of corrosion are as follows. (a) Purity of the metal: The presences of impurities in a metal accelerate its corrosion. This is because impurities form minute electrochemical cells with the metal under suitable environmental conditions, and the anodic parts get corroded. For example: Zinc metal containing impurities (such as Pb or Fe) undergoes corrosion of zinc, due to formation of local electrochemical cells. (b) Position in galvanic series: The extent of corrosion depends upon the position of metal in galvanic series. The metal or alloy which is placed at higher up in the series are more reactive and has greater tendency to undergo corrosion. The rate and severity of corrosion, depends upon the difference in their positions, and greater is the difference, the faster is the corrosion of the anodic metal alloy. (c) Over Voltage: The dissolution of metals in acids may also be considered as a corrosion reaction. Metals like Zn, Cd, Sn and Pb dissolve rather slowly in acids when they are pure. However, these metals dissolve rapidly when they contain impurities which are relatively more noble and also have a low overvoltage. The difference between the potential of the electrode (voltage) when gas evolution is actually required and expected theoretical value for the same evolution is called over voltage. For example, the presence of copper in small amounts as an impurity in zinc increases the rate of dissolution of zinc by anodic oxidation. This may be explained on the basis of the hydrogen voltage of the two metals. Pure zinc with higher hydrogen over voltage of 0.70V dissolves slowly and hydrogen evolution is also slow. Copper with low hydrogen over voltage of 0.25V also dissolves but redeposit, on the zinc surface and functions as an efficient cathode rendering the zinc anodic. Since the hydrogen over voltage is lower at the copper cathode, the rate of hydrogen evolution increases, since this cathodic reaction is favoured, the anode reaction, namely, the oxidation (corrosion) of zinc is also favoured. (d) Physical state of the metal: The rate of corrosion is influenced by physical state of the metal such as grain size, orientation of crystals, stress etc. The smaller the grain size of the metal or alloy, the greater will be its solubility in corroding medium and hence greater will be its corrosion. Moreover, areas under stress, even in a pure metal, tend to be anodic and corrosion takes place at these areas. (e) Relative areas of the anode and cathode: Rate of corrosion is less when cathodic area is less and anodic area is more. Rate of corrosion a

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6.18 Engineering Chemistry

It is clear that, due to small cathodic part, the demand for electrons will be less and this results in less dissolution of metal at the anodic part and rate of corrosion is also less. (f) Reactivity of metals: Some metals are passive in nature, Good examples are Al, Ti, Mg, Ni, Co, etc. These metals react with oxygen to form non-porous oxide layer that protects the material from further corrosion. This oxide layer on metal is also self-healing in nature, i.e., heals itself if scratched on metal surface. (g) Nature of oxide film: In aerated atmosphere, particularly all metals get covered with a thin surface film of metal oxide having a thickness of few Angstromes. Whether the metal oxide layer is protective or non-protective is decided by pilling-Bedworth rule. This rule decides the rate of corrosion in a metal. Greater the specific volume ratio, lesser is the oxidation corrosion rate eg., the specific volume ratio of Ni,Cr and W are 1.6,2.0 and 3.6 respectively, which indicates that the rate of oxidation at elevated temperature is least for Tungsten(W). (h) Solubility of the corrosion product: In electrochemical corrosion, the solubility of the corrosion products in the corroding medium is an important factor in deciding the extent and the rate of corrosion. If the corrosion product is soluble in the corroding medium, corrosion of metal will take place at a higher rate, But if the corrosion product is insoluble in the corroding medium (e.g. PbSO4 in case of Pb in a medium of H2SO4) it forms a protective layer on the metal surface and inhibits further corrosion of the metal. (i) Volatility of the corrosion product: If corrosive product is volatile in nature, they volatile as soon as they are formed. Hence, the underlying metal surface is exposed for further attack, resulting rapid and continuous corrosion.

(ii) Nature of the environment: (a) Effect of the temperature: The extent and rate of corrosion usually increases with rise in temperature. This is because an increase in temperature increases the rate of a chemical reaction as well as the rate of diffusion and decreases polarisation. (b) Effect of pH: It has been observed that the corrosion takes place more in acidic media (PH < 7) than neutral or alkaline media (PH ≥ 7). Thus, corrosion of metals can be reduced by increasing the PH of the environment contrary to it; amphoteric metals like Al, Zn and Pb are more corroded in alkaline media because they form complex ions in alkaline media and pass into solution. (c) Effect of moisture: Moisture or humidity of air is an excellent medium of corrosion. Moisture present in the atmosphere acts as a solvent for oxygen, other gases (O2, SO2 etc.) and salts and forms electrochemical cell. Hence, presence of moisture accelerates the rate of corrosion of a metal. For example, rusting of iron is quite slow in dry air but increases rapidly when the humidity of air is 60–80%. Critical humidity is the humidity of the air above which the rate of atmospheric corrosion of metal increases sharply and depends on the nature of the metal and the nature of the corrosion products. (d) Effect of corrosive gases present in air: The gases like CO2, SO2, H2S etc. present in the atmosphere or fumes of HCl, HNO3, H2SO4 etc. forms the medium more acidic above the metal surface because these gases are soluble in water to form acids and make it more conducting. This increases the rate of corrosion due to an increase in the corrosion current flowing in the miniature electrochemical cells on the metal surface.

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(e) Effect of corroding medium: Corroding medium plays an important role in deciding the rate of corrosion. Rate of corrosion is increased in the conductive corroding medium. For example, the conductance of clay and mineralised soil is much higher than those of dry sandy soils. (f) Effect of concentration of oxygen: Differential aeration concentration cell is setup due to change in the concentration of oxygen. Rate of corrosion increases with increase in concentration of oxygen. The region where oxygen concentration is lesser becomes anodic and oxygen concentration rich portion becomes cathodic. The anodic portion suffers corrosion. Rate of corrosion increases due to formation of differential aeration cell. (g) Effect of suspended particles in atmosphere: Two types of suspended particles are present in atmosphere viz., chemically active and chemically inactive. The chemically active suspended particles like NaCl. (NH4)2 SO4 absorb moisture and thus act as strong electrolytes thereby enhance corrosion rate. Whereas chemically inactive suspended particles like charcoal, absorb both moisture as well as sulphur gases and thus slowly enhances corrosion rate. (h) Effect of the nature of the presence of electrolyte: Electrolyte presence in the medium is also responsible for deciding rate of corrosion. For example, chloride ions (Cl–) present in the medium increase the rate of corrosion by destroying the passive film on metal surface; on 2− the other hand, silicate (SiO3 ) forms an insoluble layer which prevents corrosion of metal.

6.3 P ROTECTION FROM CORROSION (PREVENTIVE MEASURES FOR CORROSION CONTROL) Protection against corrosion means not allowing corrosion reactions to take place. Noble metals do not corrode but they cannot be used for common purposes, because of their high cost. We have to use other metals or alloys in the fabrication of many kinds of machinery and equipments and adopt measures to protect these from corrosion. (i) Material selection: (a) The chosen metal should be as pure as possible because the presence of impurities enhances the rate of corrosion. (b) The choice of noble metals are preferable because they are highly resistant to corrosion. (c) Avoid the contact of dissimilar metals in the presence of a corroding environment. (d) If two dissimilar metals in contact have to be used, they should be as close as possible to each other in the electrochemical series. (ii) Proper designing: (a) W hen anodic and cathodic materials are used together, then the area of anodic material should be large. (b) The anodic part should not be painted or coated because any damage in coating would cause rapid localized corrosion. (c) Whenever the direct joining of dissimilar metals, is unavoidable, an insulting fitting may be applied in-between them to avoid direct metal-metal electrical contact.

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6.20 Engineering Chemistry (d) Angles, corners, edges etc. should be avoided in construction.For this reason L, T and U shaped structures should be avoided as far as possible some better shapes of L, T and U structure are given below:

Poor-L

Better-L

Poor-T

Better-T

Water Tank

Tape

Tape Better - U

Poor - U

(e) The material should not have sharp corners and recesses because they help in accumulation of impurities. It should be avoided by proper designing as show in figure.

Weld

Recesses

Weld

Sharp Corners

Weld (Poor Design)

Sharp Corners

Weld

(Poor Design)

Weld

Smooth Bend

Weld (Best Design)

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(f) Always prevent the occurrence of in homogeneities in metal and in the corrosive environment. Thus a proper design should avoid the presence of crevices between adjacent parts of the structure, even in case of the same metal, since crevices permit concentration differences. Hence bolts and rivets should be replaced by a butt-weld as shown in figure. Crevices

Bolt joint (Poor Design)

Weld joint

(Good Design) Weld Joint

Weld joint (Best Design)

(Best Design)

(g) Whenever possible, the equipment should be supported on legs to allow free circulation of air and prevent the formation of stagnant pools or damp areas.

Air Poor design because it prevents free circulation of air

Best design because it allows free circulation of air

(h) Uniform flow of corrosion liquid is desirable, since both stagnant areas and highly turbulent flow and high velocities can cause accelerated corrosion. (iii) Cathodic protection (Electrical protection): The principle involved in this method is to force the metal to be protected to behave like a cathode, thereby corrosion does not occur. Cathodic protection is carried out by two methods: (a) Sacrificial anodic protection (Galvanic protection) In this method, the metallic structure (to be protected) is connected by a wire to a more anodic metal, so that corrosion occurs at that anodic metal and metallic structure is protected. This method is generally used for the protection of underground pipes and tanks. In this method, the more active metal like Mg is used as anode and this metal used is called as “sacrificial anode”. A block piece or plate of a more reactive metal (Zn or Mg) is buried beside the iron pipe and connected to it by a wire as shown in Figure 6.10.

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6.22 Engineering Chemistry

Soil Mg e− Magnesium anode Mg→ Mg2++2e−

iron pipe (Cathode) O2 + 2H2O + 4e− → 4OH− O2 + 4H++4e− → 2H2O

Figure 6.10 Sacrifice anodic protection Since more reactive metal (e.g., Mg) has a greater tendency to get oxidised, it undergoes oxidation in preference to iron. Thus more active metal acts as anode. At anode: Mg → Mg2+ + 2e− The electrons thus released migrate to the iron object which starts acting as cathode. At cathode released electrons reduce O2 into OH− as: At cathode: O2+ 2H2O + 4e− → 4OH− or O2+ 4H+ + 4e− → 2H2O Thus cathode (iron etc.) gets protected. Since the reactive metals (Mg, Zn etc.) scarify itself during the protection of other metal. The corroded sacrificial metal block is replaced by a fresh one, when consumed completely. Hence it is termed as sacrificial anode protection. (b) Impressed current cathodic protection In this method, an impressed current from an external source is applied in the opposite direction to neutralize the corrosion current. This is done to convert corroding metal from anode to cathode. Once the metal becomes cathodic, it is protected from corrosion. Usually, the impressed current is derived from a DC source (like battery or rectifier on a.c. line) in which negative terminal of a DC source is connected with the object to be protected is made the cathode of an electrolytic cell and positive terminal of the DC source is connected to scrap iron, platinum, graphite, nickel or lead anode (insoluble anode) and buried or immersed in a conducting medium adjacent to the metal to be protected. The anode is, usually, taken in a backfill (composed of coke, breeze or gypsum) so as to increase the electrical contact with the surrounding soil. This type of cathodic protection has been applied to protect buried structures, pipes, water-tanks etc. This process is shown in Figure 6.11.

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DC sourse

(+)

6.23

( −) Soil

Graphite Anode

e catho

Iron pip

Bockfill

Figure 6.11 Impressed current cathodic protection (iv) Surface coatings: Protecting the surface of an object by the application of coating by different methods. A brief description of two important protective coatings is given below. (a) Anodic coatings: In this process, the base metal (i.e. which is to be protected) is coated with more active metal (i.e. having lower electrode potential) such as Zn, Al and Cd coating on steel surface. If any pores, breaks or discontinuities occur in such an anodic coating, a galvanic cell is formed between the coating metal and the exposed part of the base metal, i.e. steel object. For example, in case of galvanized steel, zinc, the coating metal is attacked, leaving the underlying cathodic metal unattacked. Zinc act as anode with respect to iron, which act as cathode zinc dissolves anodically and iron metal is protected. Zinc has first corroded in the vicinity of the exposed iron spot. So, zinc coating protect iron “sacrificially”. Due to oxidation, zinc layer may be converted to basic zinc carbonate, ZnCO3. Zn (OH)2 by the action of oxygen, CO2 and moisture. This layer protects the exposed part further. This process is shown in Figure 6.12. Corrosive environment Exposed part (cathode)

Zinc Coating Zn →

Zn2+ +

−

Zn →

e− Unexposed part (anode)

Zinc Coating Zn2+ +

−

e− Flow of electrons

Unexposed part (anode)

Steel

Figure 6.12 Anodic coating i.e. galvanized steel (b) Cathodic coatings: In this process, base metal is coated with a more noble metal (i.e. having higher electrode potential). For example, coating of tin on iron, coating of copper on iron because both Sn,

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6.24 Engineering Chemistry Cu having higher electrode potential than iron. This type of coating provides effective protection to the base metal only when the base metal is completely continuous and free from pores, cracks or discontinuities. If the coating develops scratches or cracks, iron is not protected any more; the tin becomes the cathode, while the exposed iron acts as anode. This is because the standard reduction potential of iron is less than that of tin. E°Fe2+/Fe = −0.44V E°Sn2+/Sn = −0.14V A galvanic cell is set up and an intense localized attack at the small exposed part i.e. iron metal occurs, which results into severe pitting and perforation of the base metal. In such a case the rusting is much more rapid as compared to that in case of an unprotected iron piece. This process is shown in Figure 6.13. Corrosive environment Exposed part becomes anodic

Sn Coating

Fe → Fe2+ + 2e−

Unexposed part (cathode)

Flow of electrons Corrosion product

Figure 6.13 Cathodic coating i.e. Tin – plated steel Method of Application of Metal Coatings (i) Hot dipping: In this process, metal or metal alloys such as iron, copper or steel having a high melting point is coated with a low melting metals such as tin, zinc, lead or aluminum is known as hot dipping. This process involves dipping or immersing the base metal article in a molten bath of the coating metal and covered by a molten flux layer (usually ZnCl2). The flux cleans the base metal surface and prevents the oxidation of the molten coating-metal.For good adhesion; the base metal surface must be very clean; otherwise it cannot be properly wetted by the molten metal. The most commonly used hot dipping methods include (a) Galvanizing (b) Tinning or Tin plating (a) Galvanizing: The process of coating a layer of zinc on iron or steel is called galvanizing. This protects iron object from rusting. The steel article first pickled with dilute sulphuric acid to remove traces of rust, dust or any other impurities etc.; at 60–90 °C for about 15 to 20 minutes. Then the metal is dipped in a molten zinc bath at 430 °C. The surface of the bath is covered with ammonium chloride flux to prevent oxide formation on the molten zinc. When the article is taken out, it is found to have been coated with a thick layer of zinc. It is then passed through a pair of hot rollers.

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This process removes any excess of zinc and produces a thin film of uniformthickness. The coated article is annealed at a temperature of 650 °C and cooled to room temperature slowly. Galvanized articles cannot be used under acidic conditions and galvanized containers cannot be used to store acidic foods. This process is shown in Figure 6.14. Drying chamber

Iron sheet

Water Dil. H2SO4 at 60-90°C

Washing bath

Ammonium chloride flux

Pair of hot rollors

Galvanized sheet

Annealing chamber

Hot Air Molten zinc at 430°C

Excess zinc removed

Figure 6.14 Galvanisation of steel sheet (b) Tinning: The coating of tin on iron is called tin plating or tinning. In tinning, the base metal is first pickled with dilute sulphuric acid to remove surface impurities. Then it is passed through molten tin covered with zinc chloride flux. Then tin coated article is passed through a series of rollers immersed in a palm oil bath to remove the excess tin. The palm oil protects the hot tin-coated surface against oxidation. This process produces a thin film of uniform thickness on the steel sheet. Because of non-toxic nature of tin, tinning is widely used for coating steel, copper and brass sheets which is used for manufacturing containers for storing food stuffs, ghee, oils, kerosene’s and packing of food materials. This process is shown in Figure 6.15. Roller Tin-placed sheet

Pair of rollers

el Ste

she

Palm oil Bath of zinc chloride flux Molten tin Tank

Acid pickling bath

Figure 6.15 Tinning of steel sheet (ii) Metal cladding: In this process, the base metal is protecting from corrosion by coating of a thin uniform homogenous layer of a coating metal on the base metal. In this method, base metal sheet is sandwiching between thin sheets of corrosion resisting metals such as nickel, copper, lead, silver or platinum and bonded either on one side (e.g. copper clades in cooking vessels) or on both sides (e.g. duralumin is sandwiched between two layers of Pure Aluminium) permanently by the application of

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6.26 Engineering Chemistry heat and pressure. Metal cladding is generally practiced in the air craft industry in which a sheet of duralumin is sandwiched between two layers of pure aluminum to produce a sheet. The basis requirement for this specification that base metal and the cladding metal should have similar working characteristics for effective cladding. In some cases we are also used metal oxide powders in a revolving heating drum in which base metal is thoroughly immersed. This is known as diffusion or cementation of the base metal to protect from corrosion, when ZnO is used, it is known as sherardizing. When Al2O3 and Cr2O3 are used, it is known as chromizing and when only Al2O3 is used, it is known as colorizing. In all this method, we protect the base metal from corrosion by coating of thin film of different metal. (iii) Electroplating: In this process, noble metal is coated over more reactive metal. Most commonly used are tin plating and nickel plating. In electroplating, the object to be plated is made as cathode and suspended in an electroplating bath containing the metal ions to be plated by electro deposition. The anode may be of the metal to be deposited or it may be an inert electrode (such as graphite) with good electrical conductivity. During this process, the variables such as voltage, temperature, pH, current and density are kept constant so that electroplating process remains unchanged (i.e., rate of deposition of metal on cathode and rate of dissolution on anode). For example, iron can be protected from corrosion by coating the metal with chromium or nickel by electroplating process. (iv) Electro less plating: In this process, we immersed the base metal article in a bath of a noble metal salt which is used for coating. The noble metal forms a layer on the base metal article by displacement of base metal by noble metal. This process is also called as ‘immersion plating’ or ‘displacement plating.’ For example, nickel coating on base metal, In this process, base metal article is dipped in a bath of nickel sulphate and sodium hypophosphite kept at temperature of 100°C and at pH from 4.5 to 5.0. Nickel ion from solution reduces to nickel and nickel phosphide, which forms a strong adherent thin film. (v) Organic surface coatings: Organic coatings are useful for the protection of metal surface by providing inert barrier on the surface from corrosion as well as corrosive environment. Organic coating also helpful in decoration of metal surface. Organic coatings commonly used include paints, varnishes, lacquers and enamels. Paints Paint is a term which has been used to signify a uniform dispersion of finally divided solids in a liquid called “vehicle” or “medium”. The solid comprises of pigments, driers and fillers. Volatile solvent is mixed with a non-volatile forming a film on metal surface. Example of non-volatile is drying oil and volatile solvent is thinner. Constituents of paints and their function The various constituents of paint include (a) (b) (c) (d)

Pigment Vehicle or drying oil Thinner Drier

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(e) Filler or extender (f) Plasticizers and (g) Anti-skinning agent (a) Pigment: It is an essential constituent of paint. It provides color and opacity, in addition of that imparting strength and aesthetic appeal to the paint. Pigments increase the life of paint film because they prevent the penetration of UV rays which deteriorate the oil film. Many properties need to be looked for a pigment. It should be opaque, chemically inert, non-toxic and miscible with the vehicle. The pigment should have good hiding power i.e., it should be opaque so that the surface underneath is not visible; otherwise all the dirty spots, surface defects etc. would be seen. Opacity of the paint is due to the difference between the refractive indices of the pigment and the vehicle and also on the fine size of the pigment particles. Pigments commonly used in paints are inorganic solids with high refractive index either naturally occurring minerals or synthetic chemicals. Example of pigments Colour While lead [2PbCO3 ⋅ Pb(OH)2 ] White Prussian blue K 2 [Fe(CN)6 ] Blue Carbon black Black Chromium oxide (Cr2 O3 ) Green (b) Vehicle or Drying oil: It is a liquid which binds the pigment to the surface and protects pigment from decay. Common example of such oil as linseed oil, dehydrated castor oil, perilla oil or tung oil or a mixture of drying and semidrying oils. When paint is applied on a metallic surface, the unsaturated fatty acids in oil undergoes oxidation and forming oxides, peroxides and hyperoxides at the double bond and further undergo polymerisation and forming a protective, tough and insoluble film of the polymer on surface. H2C

CH2

Conjugated fatty acid residues O2 absorption decomposition

O Oxide

CH2

Non-conjugated fatty acid residues followed by rearrangement, hydrogen abstraction

O O Peroxide

Free radicals

Reactive radicals Reaction with coupling

C O O

Fatty acid components of oil and among radicals

C +

C High cross linked polymeric film

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6.28 Engineering Chemistry By adding phenolic and alkyl resin into drying oil, hardness and glossiness of the film can be improved. (c) Thinner: Thinner is a volatile solvent, which is often added to paint which helps to adjust the consistency of the paint. Other functions of thinner area (1) To increases the penetrating power of the vehicle (2) To increases the elasticity of the paint film on surface (3) It helps in retaining the constituent solids into vehicle Examples of thinness are turpentine, petroleum fractions such as benzene, naphtha, white spirit, toluol, etc. (d) Driers: Main function of a drier is to increase the drying power of the vehicle. In addition to this, driers work as oxygen-carrying catalysts which accelerates the drying of the oil film by oxidation, polymerization and condensation. Examples of common driers are borates, tungstates, resinates, linoleates of metals such as Ni, Zn, Co and Mn. (e) Filler or extender: These are often colourless inorganic substances like aluminium silicate, barium carbonate, barium sulphate, asbestos, gypsum, calcium carbonate; clay, magnesium silicate etc. are added to the paints. The function of addition of filler in paint is that it improves the properties of the paint and mainly to reduce the cost. It also acts as carriers for the pigment colour, also fill the voids in the paint film, reduce the cracking of the paint film and improve the durability of the film. (f) Plasticizer: They remain permanently in paints and varnishes. They improve the elasticity of the paint film which prevents cracking of the film. Commonly used plasticizers are tricrecyl phosphate, triphenyl phosphate, di butyl phthalate etc. (g) Anti-skinning agent: Anti-skinning agents like polyhydric phenols are added to the paint so that getting or skinning of paint can be prevented and can be used for a long period. (vi) Use of inhibitors: Inhibitors are chemical substances which on adding in small portion to the corrosive medium decreases the corrosion rate. Inhibitors are mainly of following two types (a) Anodic inhibitors: This type of inhibitors stifles the corrosion reaction, occurring at the anode by forming a sparingly soluble compound with a newly produced metal ion. Anodic inhibitors such as chromates, tungstates, phosphates of transition metal react with ions at the anode and form an insoluble precipitate. These precipitates formed are absorbed on metal surface by forming a protective film on the metal and prevent corrosion. This type of control method is effective, but it may be dangerous because if certain areas are left unprotected by depletion of the inhibitor which causes severe local attack occur on the metal surface. (b) Catholic inhibitor: This type of inhibitors slow down the corrosion reaction by considerably decreasing the diffusion of hydrated H+ ion to the cathode and can be used in acidic as well as in neutral medium. In acidic solution, the corrosion process involves the following catholic reaction. 2H+ + 2e− → H2(g) The corrosion of a metal can be reduced by slowing down the rate of diffusion of H+ ions through the cathode. It can be done by using organic compounds such as mercaptans,

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amines, substituted ureas, heavy metal soaps, heterocyclic nitrogen compounds, etc. They adsorb to the metal surface and act as cathodic inhibitors. Antimony and arsenic oxides deposit adherent film of metals at the cathode and slow down the overvoltage for hydrogen evolution. In a neutral solution, cathodic reaction is written as 1 − H 2 O + O 2 + 2e − 2OH 2 The hydroxide (OH–) ions are formed due to presence of oxygen. The corrosion can be controlled by either eliminating oxygen from the corroding environment or by retarding its movement to the cathodic areas. The oxygen is eliminated by adding reducing agents like Na2SO3 or by dearation and diffusion of oxygen to the catholic areas can be retarded by the use of Mg, Ni or Zn salts. These salts react with hydroxide ions to form corresponding insoluble hydroxides which deposit on the cathodic areas and form an almost impermeable barrier. This method is also helpful for the protection of metal surface by corrosion by slow down the corrosion process.

6.4 Review Questions 6.4.1 Fill in the Blanks 1. The gradual loss of a metal by chemical or electrochemical action of environment is called ________. [Ans.: corrosion] 2. The formula for rust is ________. [Ans.: Fe2O3 · xH2O] 3. The wet corrosion involves the flow of ________ from anodic area to cathodic area through a conducting solution. [Ans.: electrons] 4. When the oxide film is volatile in nature, rate of corrosion of underlying metal is ________. [Ans.: increases] 5. In galvanic corrosion, the metal having ________ value of reduction potential gets corroded. [Ans.: lower] 6. Larger the potential difference between two metals, ________ is the extent of corrosion. [Ans.: greater] 7. In differential aeration corrosion, the poor oxygenated part acts as ________, and undergoes corrosion. [Ans.: Anode] 8. ________ and ________ are important factors for stress corrosion. [Ans.: Tensile stress, corrosive environment]

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6.30 Engineering Chemistry 9. The rate of corrosion of a metal is inversely proportional to the ________ areas. [Ans.: anodic] 10. The rate of corrosion is more in presence of oxygen when pH value is ________. [Ans.: below 7.0] 11. The conductance of clay and mineralised soil is much higher than those of ________ soils. [Ans.: dry sandy] 12. In tinning, iron is protected with a coating of ________ metal. [Ans.: tin] 13. ________ is an example of anodic inhibitors. [Ans.: Chromates] 14. Cathodic inhibitors slow down the corrosion reaction by decreasing the diffusion of ________ ions to the cathode. [Ans.: H+] 15. In sacrificial anodic protection, the more active metal is used as ________. [Ans.: anode] 16. An example of anodic coating is ________. [Ans.: galvanization] 17. An example of cathodic coating is ________. [Ans.: tinning] 18. In metal cladding, ________ is sandwiched between two layer of pure aluminium. [Ans.: duralumin] 19. Oxidation corrosion is an example of ________. [Ans.: dry corrosion] 20. The immersion of base metal article in a bath of a noble metal salt which is used for coating is called as ________ plating. [Ans.: Electroless] 21. ________ is used to bind the pigment to the surface and protects pigment from decay. [Ans.: Vehicle/Drying oil] 22. Commonly used thinner in paint is ________. [Ans.: turpentine or petroleum] 23. Commonly used plasticizers in paint is ________. [Ans.: Tricresyl phosphate or Triphenyl phosphate] 24. Brass and copper utensils are usually coated with ________. [Ans.: Tin(Sn)]

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25. ________ are providing desired colour and protection to paint film. [Ans.: Pigments] 26. In chromizing, the base metal is thoroughly mixed in revolving drum containing metal oxides of ________ and ________. [Ans.: Al2O3, Cr2O3] 27. ________ is non-toxic in nature, so it is widely used for coating steel, copper sheets which is used for storing food stuffs and packing of food materials. [Ans.: Tinning] 28. In galvanizing, molten zinc bath is covered with ________ flux. [Ans.: Ammonium chloride] 29. ________ coating is most preferrable than ________ coating. [Ans.: Anodic, cathodic] 30. In sacrificial anodic protection, commonly ________ metal block is connected with underground pipes. [Ans.: Mg]

6.4.2 Multiple-choice Questions 1. Corrosion is an example of (a) Oxidation (c) Electrolysis [Ans.: a] 2. Chemically, the rust is (a) Fe2O3 (c) Fe2O3 · xH2O [Ans.: c]

(b) Reduction (d) Erosion

(b) FeO · Fe2O3 (d) FeO · xH2O

3. The metal which is protected by a layer of its own oxide (a) Cu (b) Fe (c) Au (d) Al [Ans.: d] 4. The corrosion caused by the direct chemical action of environmental gases or anhydrous liquids on metal surface is called (a) Dry corrosion (b) Wet corrosion (c) Pitting corrosion (d) Electrochemical corrosion [Ans.: a]

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6.32 Engineering Chemistry 5. Which of the following factors does not govern the rusting of iron? (a) Presence of air (b) Presence of moisture (c) Presence of electrolytes in water (d) Presence of impurities of more electropositive metals in iron [Ans.: d] 6. In galvanic corrosion (a) More metal gets corroded (b) Less noble metal gets corroded (c) The metal having a higher standard reduction potential gets corroded (d) The metal placed lower in the electrochemical series get corroded [Ans.: b] 7. In electrochemical corrosion (a) Anode undergoes oxidation (c) Both undergo oxidation [Ans.: a]

(b) Cathode undergoes oxidation (d) None undergoes oxidation

8. In differential aeration corrosion (a) Poor oxygenated part acts as anode (c) Poor oxygenated part acts as anode [Ans.: a]

(b) Rich oxygenated part acts as anode (d) Metal as a whole acts as cathode

9. The localised attack of a corroding environment leading to the formation of holes in an otherwise relatively unattacked surface of a metal is called (a) Water-line corrosion (b) Pitting corrosion (c) Concentration cell corrosion (d) Wet corrosion [Ans.: b] 10. Water-line corrosion is enhanced by the presence of (a) Hydroxides (b) Chlorides (c) Carbonates (d) Silicates [Ans.: b] 11. Caustic embrittlement is a particular case of (a) Pitting corrosion (b) Dry corrosion (c) Stress corrosion (d) Wet corrosion [Ans.: c] 12. To protect buried pipeline from corrosion is connected to Mg piece through a wire. This process is called as (a) Impressed current cathodic protection (b) Galvanic protection (c) Sacrificial anodic protection (d) Sacrificial cathodic protection [Ans.: c]

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13. In an electrochemical series, the metal at the top is (a) Most noble (b) Most stable (c) Most active (d) Most protective [Ans.: c] 14. Galvanizing is the process of coating iron with (a) Mg (b) Cu (c) Zn (d) Ni [Ans.: c] 15. Corrosion of zinc metal containing an impurity of copper is called (a) Water line corrosion (b) Moist corrosion (c) Stress corrosion (d) Galvanic corrosion [Ans.: d] 16. Anodic coating protects underlined metal (a) Due to its higher reduction potential (b) Due to its lower reduction potential (c) Due to its noble nature (d) Due to its higher oxidation potential [Ans.: d] 17. Addition of hydrazine-hydrate to corrosive environment (a) Retard anodic reaction (b) Retard cathodic reaction by consuming dissolved oxygen (c) Prevents diffusion of protons to cathode (d) Increases hydrogen overvoltage [Ans.: b] 18. In general, corrosion is maximum when the pH of the corroding medium is (a) Above 7.0 (b) Equal to 7.0 (c) Below 7.0 (d) Equal to 1.0 [Ans.: c] 19. The process of covering steel with zinc to prevent it from corrosion is called (a) Galvanizing (b) Tinning (c) Electroplating (d) Electroless plating [Ans.: a] 20. Acid pickling of steel is carried out by dipping the steel in (a) Dilute Hcl (b) Dilute H2SO4 (c) Conc H2SO4 (d) Dil HNO3 [Ans.: b]

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6.34 Engineering Chemistry 21. During galvanization, the function of flux ammonium chloride is (a) To prevent oxide formation, on molten zinc (b) To prevent reduction of molten zinc (c) To acts as a barrier (d) None of these [Ans.: a] 22. In electroplating, the object to be protected from corrosion is made as (a) Anode (b) Cathode (c) Both anode and cathode [Ans.: b] 23. The oxygen carrier of the paint is called (a) Drier (c) Thinner [Ans.: a]

(d) None of the above

(b) Pigment (d) Extenders

24. In Electroless plating, the base metal article is immersed in a solution of (a) More active metal salt (b) More noble metal salt (c) Any one of these (d) None of the above [Ans.: b] 25. An inhibitor which when added in small quantities to aqueous corrosive environment (a) Effectively decreases the corrosion of a metal (b) Increases the corrosion of a metal (c) No effect on corrosion of metal (d) Increases the corrosion nature of the environment [Ans.: a] 26. The cathodic inhibitors slow down the corrosion reaction by decreases (a) Diffusion of hydrated H+ ion to the cathode (b) Diffusion of cl- ions to the cathode (c) Diffusion of hydrated H+ ion to the anode (d) None of the above [Ans.: a] 27. In cathodic coating, base metal is coated with (a) More noble metal (b) Less noble metal (c) More active metal (d) Having more reduction potential [Ans.: a]

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28. In Impressed current cathodic protection, anode is placed in backfill because (a) To slow down the rate of corrosion reaction (b) To increases the rate of reaction (c) To increase the electrical contact with the surrounding soil (d) None of the above [Ans.: c] 29. The rate of corrosion is more when (a) Anodic area is large (b) Anodic area is small (c) Athodic area is small (d) None of the above [Ans.: b] 30. According to pilling-Bedworth rule, Greater is the specific volume ratio (a) More is the oxidation corrosion (b) Lesser is the oxidation corrosion (c) More is the reduction corrosion (d) None of the above [Ans.: b]

6.4.3 Short Answer Questions 1. Define corrosion. Ans.: Any process of deterioration and consequent loss of solid metallic materials through an unwanted chemical or electrochemical attack by its environment, is called as corrosion. 2. What is meant by rusting of iron. Ans.: The attack of atmospheric gases on iron or steel, formation of a layer of reddish scale of hydrated ferric oxide fe2O3 · 3H2O on its surface is known as rusting of iron. 3. What is dry corrosion. Ans.: Dry corrosion takes place due to the direct chemical action of atmospheric gases like CO2, SO2, O2, H2 etc or anhydrous liquids on the metal surfaces. 4. Formation of which types of metal oxide film cause rapid and continuous corrosion. Ans.: Volatile oxide film and porous oxide film. 5. Formation of which types of metal oxide film prevents corrosion. Ans.: Highly unstable oxide film and finely grained tightly adhering, impervious oxide film. 6. State the two conditions for wet corrosion to take place. Ans.: (i) Immersion or partial dipping of two dissimilar metals or alloy in a solution. (ii) A metal in contact with the conducting liquid.

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6.36 Engineering Chemistry 7. Bolt and nut made of the same metal is preferred in practice. Why? Ans.: Because such a combination will not permit galvanic corrosion to take place. 8. What is wet corrosion. Ans.: Wet corrosion is due to the flow of electrons from metal surface anodic area towards cathodic area through a conducting solution. It is also known as electrochemical corrosion. 9. What is galvanic corrosion. Ans.: W hen two dissimilar metals are electrically connected and exposed to an electrolyte, the metal higher in electrochemical series undergo corrosion. 10. The rate of metallic corrosion increases with increase in temperature. Give reason. Ans.: With increase of temperature of the environment, the rate of reaction as well as rate of diffusion increases, thereby corrosion rate increases. 11. Iron corrodes faster than aluminium, even though iron is placed below aluminium in the electrochemical series, why? Ans.: This can be explained by the fact that aluminium forms a non-porous, very thin, highly adhering protective oxide film (Al2O3) on its surface and this film does not permit corrosion to occur. 12. Wire mesh corrodes faster at the joints, why? Ans.: The joints of wire mesh are stressed due to welding, so that part acts as anode, Hence oxidation takes place easily at such joints leading to faster corrosion at the joints of wire mesh. 13. Impure metal corrodes faster than pure metal under identical conditions. Why? Ans.: Because presence of impurities in metal cause heterogeneity and form minute electrochemical cells at the exposed parts, and anodic parts get easily corroded. 14. How is galvanization different from cathodic protection. Ans.: In galvanization, the iron object is protected from corrosion by coating it with a layer of zinc, whereas in cathodic protection, the iron object in made cathodic by connecting it with a more anodic metal like Al, Mg, etc. 15. Where the electrochemical corrosion takes place. Ans.: At the anodic area. 16. Rusting of iron is faster in saline water than in ordinary water. Give reason. Ans.: Due to presence of sodium chloride in saline water, it leads to increased conductivity of water, so when saline water comes in contact with the iron surface, corrosion current increases and rusting is speeded up. 17. Why does part of a nail inside the wood undergoes corrosion easily? Ans.: Corrosion is due to differential aeration, Because part of nail inside the wood is not exposed to atmospheric conditions, whereas the nail outside is exposed to atmospheric air. Thus nail inside the wood becomes anodic while remaining part acts as cathodic. So due to differential aeration, a differential current starts flowing, and the anodic parts gets corroded easily.

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18. Why should nickel plated steel articles be free from pores and pin holes. Ans.: with respect to nickel, steel is anodic and if there are pin holes and pores in nickel plated steel article, they will expose the anodic steel to atmosphere. A galvanic cell is set up and an intense localized corrosion at these small exposed parts occur. 19. Corrosion of water filled steel tanks occurs below the water line. Why? Ans.: This is because, the area above the waterline is highly oxygenated and acts as cathodic, while the part below the waterline is poorly oxygenated and acts as anodic. So due to differential aeration, an electrochemical cell is set up which result in corrosion of steel tanks below the waterline. 20. What is meant by the term passivity? Ans.: It is the phenomenon by which a metal or alloy shows higher corrosion resistance due to formation of a highly protective, very thin and quite invisible surface film on metal surface. 21. What is effect of pH on corrosion. Ans.: The lower the pH (or more acidic), greater is the corrosion. 22. Can we use aluminium in place of zinc for cathodic protection of rusting of iron, comment. Ans.: Standard electrode potential of Al3+/Al = -1.66V

Zn2+/Zn = -0.76V In cathodic protection, the metal (iron) to be protected from corrosion is connected by a wire to a more anodic metal (like Al, Zn etc.), so that all the corrosion occurs at this more active metal. Thus, the parent metal is protected while the more active metal gets corroded slowly. As the standard potential of aluminium is more than zinc, so Al is more anodic than Zn, so we can better use aluminium in place of zinc for cathodic protection of rusting of iron.

23. Why are galvanized utensils not used? Ans.: Because galvanized articles gets dissolved in dilute food acids and forms highly toxic compounds. So, galvanized utensils cannot be used for preparation and storing food stuffs. 24. Why are brass utensils usually tinned? Ans.: Because Tin (Sn) is a noble metal and protects the brass utensils from corrosion, moreover, tin is non-toxic in nature. Hence, it is widely used for coating copper and brass utensils. 25. Galvanization of iron article is preferred to tinning, why? Ans.: Galvanization (coating iron with zinc) is preferred to tinning (coating iron with tin) due to the following reason: zinc(Zn) is more electropositive than iron, so zinc coating acts as anode; while the exposed iron portions of coating acts as cathode, If by chance, the zinc coating is broken at some place, the zinc (being more anodic than iron), undergoes corrosion, protecting iron from rusting. So, zinc coating protects iron sacrifically. On the other hand, tin is a noble metal (i.e. having higher reduction potential than iron), so it protects the iron due to its higher corrosion resistance than iron, If by chance, the tin coating is broken at some place, much more corrosion of iron takes place because small exposed part of iron cuts as anode and tin acts as cathode, a galvanic cell is set up, thereby an intense corrosion at the small exposed iron part occurs.

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6.38 Engineering Chemistry 26. What is chromizing? Ans.: The process of heating a mixture of chromium powder, alumina and iron/steel article in a revolving drum at 1300-1400 °c for 3-4 hours, It increases the corrosion resistance of the article. 27. What are the main constituents of oil varnish. Ans.: synthetic resin, drying oil and volatile solvent 28. Give two functions of plasticizers. Ans.: (i) It provides elasticity to the paint film. (ii) To minimize the cracking of dried paint film. 29. Give three functions of drier in paints. Ans.: (i) It acts as a carrier of pigments (ii) It helps in forming a thin, homogeneous and protective film. (iii) It supplies to paint film adhesion, toughness, durability and water-proofness. 30. Give two functions of extenders or fillers Ans.: (i) It reduce the cost of paint. (ii) It reduce the cracking of the paint film. 31. What is an enamel? Ans.: enamel is an intimate dispersion of pigment in a varnish. 32. Give three function of thinner in a paint. (i) To suspend pigment particles (ii) To dissolve film-forming materials. (iii) To reduce the consistency of paint for getting smooth finish.

6.4.4 Descriptive Questions Q.1 Define corrosion of metals. Explain the electrochemical theory of wet corrosion with mechanism. Q.2 Give reasons for the following: Silver and copper do not undergo much corrosion like iron in moist atmosphere. Q.3 Write short notes on the following: (i) Wet corrosion (ii) Dry corrosion (iii) pitting corrosion Q.4 Explain the mechanism of galvanic corrosion and differential aeration corrosion. Q.5 What are the factors affecting corrosion? How is it prevented? Q.6 Explain how can corrosion be controlled by proper designing. Q.7 What is the role of sacrificial anode in corrosion control. Q.8 What are the effects of temperature, pH, overvoltage and reactivity of metal influences the corrosion.

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Q.9 Write short notes on the following: (a) Sacrificial anodic protection (Galvanic protection) (b) Impressed current cathodic protection (c) Galvanization (d) Galvanic series (e) Caustic embrittlement Q.10 Discuss the role of nature of oxide formed in oxidation corrosion. State and explain pillingBedworth rule. Q.11 Explain the mechanism of hydrogen evolution and oxygen absorption in electrochemical corrosion. Q.12 Describe the following methods of corrosion control (i) Tinning (ii) hot dipping (iii) proper designing (iv) electroplating (v) metal clading. Q.13 What is meant by corrosion inhibitors. Give two examples. Q.14 Explain the importance of tinning in corrosion control. Q.15 Give reasons for the following: (i) corrosion of water filled tank occurs below the waterline. (ii) A copper equipment should not possess a small steel bolt. Q.16 Discuss the importance of design and material selection in controlling corrosion. Q.17 Explain (i) pitting corrosion (ii) Bi-metallic (Galvanic) corrosion. Q.18 Outline the difference in the use of anodic and cathodic coatings for corrosion prevention. Q.19 Describe the process of galvanization of iron. How does it prevents the corrosion of iron. Q.20 (i) Give the requirements of a good paint. (ii) Write brief account on pigments. Q.21 Explain the principle involved in (i) anodic protection (ii) Cathodic protection (iii) galvanization (iv) cementation Q.22 What happens and why? (i) Iron sheet riveted with copper rivets. (ii) An iron pole is partly burried under earth. (iii) Zinc plate fixed below the ship. Q.23 Iron corrodes faster than aluminium, even though iron is placed below aluminium in the electrochemical series, why? Q.24 Explain rusting of iron with the help of electrochemical theory of corrosion. Q.25 Discuss the differences between varnishes and paints.

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6.40 Engineering Chemistry Q.26 What are the constituents of paints and what are their function. Q.27 What are the important factors that influence the corrosion phenomenon. Q.28 Write short notes on (i) Sacrificial anode (ii) Corrosion inhibitors (iii) Electroplating (iv) Electroless Plating Q.29 How does the nature of metal influence rate of corrosion. Q.30 How are the metals protected against corrosion by modifying the environment?

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Chemistry of Engineering Materials

7.1 SEMICONDUCTING AND SUPER CONDUCTING MATERIALS According to conductivity, elements or materials are broadly classified into three categories. They are as follows: (i) Conductors (ii) Semiconductors (iii) Insulators Conductors They allow the maximum portion of the applied thermal or electric field to flow through them. For example, metals are good conductors. Insulators They do not practically allow the heat or electricity to flow through them. For example, most organic and inorganic solids, except graphite. Semiconductors The thermal and electrical conductivity of a semiconductor at normal temperature lies between that of a conductor and an insulator. Semiconductors are those solids which are perfect insulators at absolute zero, but conduct electric current at room temperature. For example, silicon and germanium are two important elements used as semiconductors.

7.1.1 Semiconductor At room temperature, semiconductors allow a portion of electric current to flow through them. The electrical conductivity of a semiconductor at normal temperature lies between that of a good conductor and an insulator in the range of 10-9 to 102 ohm-1cm-1.

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7.2 Engineering Chemistry Semiconductors are those solids which are perfect insulators at absolute zero, but conduct electric current at room temperature. Silicon and germanium are two very important elements used as semiconductors. The pure samples (≥99.999% pure) of these elements are obtained by zone refining; some impurity is added deliberately to these elements by a process called doping. These are of two types— intrinsic and extrinsic. Intrinsic Semiconductors (Semiconductors Due to Thermal Defects) Pure silicon or germanium act as insulators because electrons are fixed in covalent bonds and are not available for conduction. At high temperatures, electrons are released by breaking of some of the covalent bonds. These electrons move freely in the crystal and can conduct electricity. Without introducing an external substance, these materials show conduction. Hence, these materials are known as intrinsic semiconductors. Extrinsic Semiconductors (Semiconductors Due to Impurity Defects) Silicon and germanium (group 14 elements) in pure state have very low electrical conductivity. However, the low electrical conductivity of these elements is greatly enhanced by the addition of even traces of an element belonging to groups 13 (iii) or group 15 (v) to the crystals of group 14 (iv) elements, that is, silicon or germanium. The induction of group 15 and group 13 elements to the crystal lattice of group 14 elements (Si or Ge) produces n-type semiconductors and p-type semiconductors respectively. n-type Semiconductors ( n Stands for Negative) This type of semiconductor is produced in any of the following ways: (i) Due to metal excess defect (explained earlier) (ii) By the addition of trace amount of group 15 element (P, As) to extremely pure germanium or silicon by a process called doping. When an element of group 15(As) is added to germanium (group 14 element) crystal, some atoms of germanium are replaced by arsenic. In such cases, four electrons of the impurity element (As) are used in forming bonds to Ge, while the fifth electron remains unused. The additional electrons can move freely and conduct electricity in the metals. Hence, arsenic-doped germanium exhibits fairly high electrical conductivity. p-type Semiconductor ( p Stands for Positive) This type of semiconductors are formed by the following ways: (i) Due to metal deficiency defects (ii) By addition of impurity atoms containing less electrons (i.e., atoms of group 13); in such cases, the parent insulator to the insulator lattice behaves like a conductor–electron. When a group 13 element (like B, Ga, In) substitutes for a germanium atom (group 14 element), that is, when a group 13 element (say B, having only three electrons in the outer shell) is added in small traces to group 14 elements (say germanium), the atoms of B are not able to complete tetrahedral covalent structures because they have one electron short of the requirement. Hence, some of the sites normally occupied by electrons will be left empty. This gives rise to electron vacancies, commonly known as

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positive holes because the net charge at these sites is positive. With an electric field, the adjacent electrons move into the positive holes and form other positive holes. The current is passed in the crystal due to the migration of positive holes. The electrical conductivity of germanium (group 14 element) crystal increases by the doping trace amount of B (group 13 element). Here, the current is carried by positive holes; hence, this type of conduction is known as p-type semi conduction. Unlike metals, the conductivity of semiconductors increases with increase in temperature. This is due to the fact that an extra electron or a positive hole is weakly bound with the crystal; when energy is supplied in the form of heat, they (electron or a positive hole) become force from the crystal lattice for the conduction of electricity.

7.1.2 Applications of Semiconductors Various semiconductors have been prepared by the following types of combination. (i) (ii) (iii) (iv)

Elements of group 14 (Se, Ge) and group 15 (P, As, Sb) Elements of group 13 (B, Ga) and group 14 (Si, Ge) Elements of group 13 and group 15 (In Sb, Al, P) Elements of group 12 and group 16 (ZnS, CdS, CdSe, Hg, Te)

The properties of a semiconductor are considerably changed depending upon the nature of the impurity. Semiconductors are used in transistors and in exposure metals as photoelectric devices. A combination of p- and n-type of semiconductors (known as p-n junction) allows electric current from outside to flow through it in one direction. This type of p-n junction is known as a rectifier and is used for converting alternating current to direct current.

7.1.3 Superconductors The electrical resistance of metals depends upon temperature. The electrical resistance decreases with a decrease in temperature and becomes almost zero near the absolute temperature. Materials in this state are said to possess super-conductivity. Thus, super-conductivity may be defined as a phenomenon in which metals, alloys and chemical compounds become perfect conductors with zero resistivity at temperatures approaching absolute zero. Super-conductors are diamagnetic. The phenomenon was first discovered by Kamerlingh Onnes in 1913 when he found that mercury becomes super-conductor at 4K. The temperature at which a substance behaves as super-conductor is called transition temperature. Most metals exhibit this phenomenon at a temperature range of 2K and 5K. Efforts are being made to find materials that behave as super-conductors at room temperature because attaining low temperature with liquid helium is highly expensive. The highest temperature at which super-conductivity has been observed is 23K for alloys of niobium (Nb3Ge). Since 1987, many complex metal oxides have been found to possess super-conductivity at fairly high temperatures. Some examples are given here. YBa2Cu3O7 90K Bi2Ca2Sr2Cu3O10 105K Tl2Ca2Ba2Cu3O10 125K Super-conductors have many application in electronics, building magnets, aviation transportation (trains which move in air without rails) and power transmission.

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7.4 Engineering Chemistry

7.2 MAGNETIC MATERIALS Magnetism is the ability of matter in which there is a force of attraction or repulsion between unlike or like poles. More than 2,000 years ago, ancient Greeks discovered a mineral that attracts things made of iron. This mineral was found in Magnesia, a part of Turkey; hence it was named magnetite. Magnets are commonly used in workplaces and homes.

7.2.1 General Properties of Magnetic Materials Some of the properties of magnetic materials are as follows: (i) The Earth acts as a big bar magnet through its core. The north pole of magnets and compass needles point to the Earth’s magnetic south pole, which is near the Earth’s geographic north pole. (ii) All magnets have two poles. If a magnet is allowed to rotate freely, the north pole will always point to the north and the other is called the south pole. (iii) Opposite magnetic poles attract each other and like magnetic poles repel. (iv) Every magnet is surrounded by a magnetic field. The magnetic field lines explain the shape of the field. (v) According to Becquerel and Faraday, all matter including liquids and gases were affected by magnetism, but a few respond to a noticeable extent and others do not.

7.2.2 Classification of Magnetic Materials According to Faraday’s law of magnetic induction, the magnetic forces of material electrons will be affected when a material is placed within a magnetic field. However, in a magnetic field material, electrons can react quite differently (attract/repel). This mainly depends on the atomic or molecular structure of the material and the net magnetic field associated with the atom. Depending on the attraction and repulsion in the magnetic field, materials can be classified into five categories as follows. (i) (ii) (iii) (iv) (v)

Diamagnetic materials—weak repulsion to field Paramagnetic materials—weak attraction to field Ferromagnetic materials—strong attraction to field Ferrimagnetic materials—strong attraction to field Antiferromagnetic materials—no magnetic moment

Diamagnetic Materials Diamagnetic materials have weak magnetic susceptibility; hence, they repel slightly in a magnetic field and the material does not retain the magnetic properties when the external field is removed. Such materials have no permanent net magnetic movement due to paired electrons. In the external magnetic field, the electron paths are realigned; hence, the material shows weak repulsion. For example, copper, silver, gold, etc. Paramagnetic Materials Paramagnetic materials have a small positive susceptibility; hence, they attract slightly in a magnetic field and the material does not retain the magnetic properties when the external field is removed.

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Due to the presence of some unpaired electrons and the realignment of the electron path with the external magnetic field, the materials show paramagnetism. For example, magnesium, molybdenum, lithium, tantalum, etc. Ferromagnetic Materials Ferromagnetic materials have large positive susceptibility; hence, they exhibit strong attraction to magnetic field and the materials retain their magnetic properties after the external field has been removed due to the presence of magnetic domains. Due to the presence of unpaired electrons, ferromagnetic materials have net magnetic moment. Here, all the magnetic dipoles are aligned parallel and are oriented in the same direction. For example, iron, nickel, cobalt, etc. Curie Temperature At a particular temperature, the electronic exchange forces in ferromagnets are very large. Hence, thermal energy eventually overcomes the exchange and produces a randomising effect, and that temperature is known as “Curie temperature”. Below the Curie temperature, the ferromagnet is ordered and the above is disordered. Hysteresis Retaining magnetic properties after the removal of external magnetic field is known as hysteresis. Magnetic Domain In ferro and ferrimagnetic materials below Curie temperature, a large number of atom moments is aligned parallel as small volume regions. This is known as “domain”. The adjacent domains are separated by boundaries and are shown in Figure 7.1.

Domain boundary Another domain

One domain

Figure 7.1 Adjacent domains separated by boundaries Ferrimagnetic Materials Ferrimagnetism is similar to ferromagnetism, but is observed in complex crystals and not in atoms in which the magnetic moments of neighbouring ions are antiparallel and unequal in magnitude. For example, magnatite was considered a ferromagnet until 1940. In 1940, Neel provided the theoretical framework about ferrimagnetism.

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7.6 Engineering Chemistry Antiferromagnetic Materials Antiferromagnetism is a phenomenon exhibited by materials in which the complete magnetic movement is cancelled with the antiparallel coupling of adjacent atoms or ions. Here, the successive magnetic dipoles are aligned in opposite directions with the same magnitude; hence, it has no net magnetic moment. For example, manganous oxide, chromium, etc. Magnetic materials and their spin alignment are shown in Figure 7.2. Paramagnetism

Ferromagnetism

Antiferromagnetism

Ferrimagnetism

Figure 7.2 Magnetic materials and spin alignment

7.2.3 Applications of Magnetic Materials Magnets are used in a vast array of products from loudspeakers to space research. Some applications are as follows: (i) Power conversion (electrical to mechanical): In motors (starter motor, power steering motors, washer pumps), generators and electromagnets. (ii) Power adaptation and signal transfer: Transformers. (iii) Permanent magnets: Loudspeaker, sensors, navigation and information systems. (iv) Data storage analogue: Video and audio tapes. (v) Data storage digital: Hard disk, floppy disk. (vi) Quantum devices: Magnetoresisstive random access memory (MRAM), Gaint magnetoresistance (GMR) reading head. (vii) Instrumentation: Dashboard instruments, nanoscience and technology, medicine, research, etc.

7.3 CEMENT Concrete is a widely used non-metallic material in construction. Cement is an important bonding material and can bond sand and rock with water in concrete. It has adhesive and cohesive nature and can bond with bricks, stones, etc.

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7.3.1 Classification of Cement Cement is broadly classified into natural, puzzolana, slag and Portland cement. These are briefly discussed here. Natural Cement Natural cement is prepared with the calcination and pulverisation of naturally occurring argillaceous limestone at high temperature. During calcination, calcium silicates and aluminates are formed. Natural cement is a setting cement and possesses hydraulic qualities and relatively low strength. The combination of sand with natural cement is known as mortar and is used in laying bricks and setting stones. Mortar is also used in the construction of dams and as a foundation for bulk masses of concrete. Puzzolana Cement Puzzolana cement is the oldest cement. It is invented by Romans and is used by them for the construction of walls and domes. The mixing and grinding of natural puzzolana and slaked lime gives puzzolana cement. Volcanic ash produced by rapid cooling of lava is known as natural puzzolana. It is a molten mixture of silicates of calcium, aluminium and iron and has hydraulic properties as well. Slag Cement Slag cement is a mixture of lime and blast furnace slag. A mixture of calcium and aluminium silicates (blast furnace slag) is granulated by pouring into a stream of cold water, dried mixed with hydrated lime and pulverised to fine powder. The setting of slag cement is too slow, poor in abrasion resistance and lower in strength. Hence, slag cement has limited applications and is used for making concrete in bulk construction. Portland Cement In 1824, William Aspdin prepared Portland cement by heating limestone and clay by crushing the resulting product to a fine powder. Hence, he is known as father of Portland cement. On mixing with water, the cement is set to give a hard stone-like mass and resembles the stone of Portland, England. Hence, it is known as a magic powder.

7.3.2 Raw Materials used in the Manufacture of Portland Cement Portland cement primarily consists of lime, silica, alumina and iron. The following materials are used for manufacture of cement: (i) Calcareous materials, CaO (such as limestone, chalk, marl, etc.) (ii) Argillaceous materials, Al2O3 (such as clay, shale, aluminium ore refuse, fly ash and so on) (iii) Siliceous material, SiO2 (such as clay, shale, sand and so on) (iv) Powdered coal or fuel oil

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7.8 Engineering Chemistry (v) Gypsum (CaSO4 · 2H2O) and (vi) Iron components, Fe2O3 (ferriferous materials such as clay, iron ore and so on) Importance and the Effects of the Ingredients on Cement The proportion of the ingredients should be properly maintained; otherwise, the following effects may be observed on the characteristics of cement. Strength: Lime, silica, iron oxide and alumina play a vital role on strength of the cement. However, excess/loss amount of lime and alumina reduces the strength due to expansion and disintegration. Colour and hardness: Iron oxide provides colour and hardness to cement. Soundness: A small amount of sulphur trioxide imparts soundness to cement; however, excess amount reduces the soundness. Setting: Alumina and lime help in quick setting. Gypsum helps to retard the setting action of cement and enhances the initial setting time. Efflorescent: Excess alkali causes the efflorescent.

7.3.3 Manufacture of Portland Cement The manufacture of Portland cement involves the following steps: (i) Crushing (ii) Mixing (iii) Burning (iv) Grinding (v) Packing Crushing Crushing of raw materials is done with two crushers. The primary crusher reduces the size of the raw material to approximately five inches and the secondary crusher further reduces the size to threefourth inches. These are then ground to a fine powder (in ball mills or tube mills). Each separate powdered ingredient is stored in separate hoopers (Figure 7.3). Drilling RIG Overburden Shale

To crusher n

tin

a ibr

v Limestone To Raw materials consists of combinations of limestone, cement rock, mart or oyster shells, Primary crusher and shale, clay, sand, or iron ore

ee cr

Each raw material is stored separately

Secondary crusher

Raw materials conveyed to grinding mills

Figure 7.3 Stone being reduced to five inches and three-fourths and stored

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Mixing Mixing of raw materials can be done either by a dry or a wet process. Dry Process The following proportions of the powdered materials, that is, lime 60–69 per cent, silica 17–25 per cent, alumina 3–8 per cent, iron oxide 2–4 per cent, magnesium oxide 1–5 per cent, alkali oxides like Na2O + K 2O 0.3–1.5 per cent and sulphur trioxide 1–3 per cent are then mixed and we get a raw mix. This is stored in silos (storage bins) and are sent to a rotary kiln for burning. This process is shown in Figure 7.4 (a) and (b).

Raw mix

Iron ore

Dust collector Hot air furnace

Oversize Fines

Clay

Limestone

Cement rock

Dry process

To air separator

Raw materials are propertioned

Grinding mill

To pneumatic pump

To Kiln

Dry mixing and blending silos

Ground raw material storage

Raw materials are propertioned

Oversiz

Grinding mill

Fines

Water added here

Vibrating screen Slurry

Iron ore

Clay

Limestone

Cement rock

Wet process

(a)

Slurry pump

To Kiln

Slurry is mixed and blended

Slurry Storage basins pump

Raw materials are ground to power and blended (b)

Figure 7.4 (a) Dry process: mixing of raw materials; (b) Wet process: mixing of raw materials with water Wet Process The calcareous raw material is crushed, powdered, stored and the argillaceous material (clay) is thoroughly mixed with water for removing organic material in wash mills stored in basins. These two are led to grinding mills (tube mill/ball mill) through channels in the right proportions and are mixed to form a paste called slurry. The chemical composition of slurry may be adjusted with correcting basins, and it contains about 38–40 per cent of water and is stored in tanks for feeding to a rotary kiln. Differences between dry and wet process of mixing raw materials shown in Table 7.1.

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7.10 Engineering Chemistry Table 7.1 Differences between dry and wet processes Dry process

Wet process

1. It is used only when the raw materials are too hard. 2. In this method, fine ground powder is formed. 3. This process is slow and fuel consumption is low. 4. Inferior quality cement is formed. 5. The cost of production is low but it is highly economical.

It can be used for any kind of raw material. Here, slurry is formed. This process is fast and fuel consumption is high. Superior quality cement is formed. Cost of production is little high, but the process is cheap.

Burning The burning process is done in a rotary kiln containing a steel tube, lined inside with refractory bricks, having 2.5–3m in diameter and 90–120m in length. The kiln is in a slightly inclined position and is capable of rotating at 1 rpm along its longitudinal axis. Fuel and air are injected at the lower end for burning, which produce long hot flames that heat the interior of the kiln up to 1,750°C. From the upper end of the kiln, raw mix or corrected slurry is injected. From the lower end of the kiln, hot flames are forced with slow rotation and through slope of the kiln, the fed material move towards the bottom of the kiln and the material descends gradually with temperature. Depending on the temperature, the kiln is divided into three zones. They are drying zone, calcination zone and clinkering zone. Drying zone: It is the upper part of the kiln having temperature around 400°C. Here, water in the slurry gets evaporated. Calcination zone: It is the central part of the kiln having temperature around 1,000°C. Here, limestone of the dry mix or slurry is decomposed to give quick lime as small lumps, also called modules and carbon dioxide, escape out. CaCO3

Limestone

CaO + CO2 ↑

Quicklime

Clinkering zone: It is the lower part of the kiln having temperature between 1500 and 1700°C. Here, the chemical interaction of fusion occurs between lime and clay to form calcium aluminates and silicates. 2Cao + SiO2 → Ca 2SiO 4 (C2S) Dicalcium silicate

3Cao + SiO 2 → Ca 3SiO5 (C3S)

Tricalcium silicate

3Cao + Al 2 O3 → Ca 3 Al 2 O6 (C3 A)

Tricalcium aluminate

4Cao + Al 2 O3 + Fe2 O3 → Ca 4 Al 2 FeO10 (C 4 AF) Tetra calciuminoferrite

The silicates and aluminates of calcium fuse to form about 0.5–1cm diameter hard, greyish stones, known as clinkers. These are hot at about 1,000°C. They are cooled with cool air in another small rotary kiln at the base of the main kiln and are collected in small trolleys (Figure 7.5).

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Chemistry of Engineering Materials

Dust Collector

To kiln

Fan

Materials are stored separately Clinker

Raw mix is kiln burned To partial fusion at 2700 °F

Coal oil or gas fuel

Rotating kiln

Dust bin

7.11

Clinker cooler

Gypsum

Clinker and gypsum conveyed to grinding mills

Figure 7.5 Burning changes raw mix chemically into cement clinker Grinding In ball mills or tube mills, the cooled clinkers are ground to fine powder and 2–3 per cent of gypsum is added to avoid quick setting and also acts as a retarding agent for early setting of cement, this is shown in Figure 7.6. 3CaOAl 2 O3

Tricalcium aluminate after initial set

Gypsum

Clinker

Air separator

Materials are proportioned

rsize

Ove

xCaSO 4 7H 2 O

→

Gypsum

3CaO ⋅ Al 2 O3 + xCaSO 4 ⋅ 7H 2 O Insoluble tricalcium sulphoalluminate

Dust collector

Fines

Grinding mill

Cement pump

Bulk storage

Bulk Bulk Box Packaging Truck truck car car machine

Figure 7.6 Grinding—clinker with gypsum added into Portland cement and shipped After the initial set, the cement water paste becomes stiff, but gypsum retards the dissolution of C3A by forming tricalcium sulphoaluminate (3CaO ⋅ Al 2 O3 ⋅ x ⋅ CaSO 4 + H 2 O), which is insoluble (Figure 7.6). Thus, 3CaOAl 2 O3 + xCaSO 4 ⋅ 7H 2 O → 3CaO ⋅ Al 2 O3 xCaSO 4 ⋅ 7H 2 O After initial set

Gypsum

Tricalcium sulpho aluminate (insoluble)

The formation of insoluble C3A prevents very early further reactions of setting and hardening. Packing The ground cement is stored in silos, from which it is fed to automatic packing machines. Each bag, usually contains 50kg of cement. The flow diagram for the manufacture of Portland cement is shown in Figure 7.7.

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7.12 Engineering Chemistry

7.3.4 Chemical Composition of Portland Cement and its Importance Portland cement mainly contains dicalcium silicate (C2S), tricalcium silicate (C3S), tricalcium aluminate (C3A) and tetracalcium alumina ferrite (C4AFe). Each component exhibits particular special behaviour, hence the behaviour of cement can be altered by changing the relative percentages of the aforementioned compounds. Dicalcium Silicate (C2S) Due to slow reaction with water, it gets hardened slowly and strengthens after one week with the formation of tobermonite gel with high surface area. Moist curing continues up to six months. 2CaO ⋅ SiO2 + xH 2 O → 2CaO ⋅ SiO2 ⋅ xH 2 O Dicalcium silicate

Tobermonite gel

Raw-materials Limestone

Clay

1. Crushing

Washing Water

Proportioning Grinding 2. Mixing

(Dry process)

(Wet process)

Dry powder

Slurry

Water

3. Burning

Rotary kiln

Pulverised coal

Coal

Clinkers 4. Grinding Fine clinkers Gypsum Portland cement 5. Packaging Shipment

Figure 7.7 Flow diagram for the manufacture of Portland cement

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Tricalcium Silicate (C3S) This material is responsible for the initial setting and early strength by the formation of hydrolgel; it has a binding action between the aggregates when rapid reaction occurs with water. 3CaO ⋅ SiO2 + xH 2 O → Ca(OH)2 + 2CaO ⋅ SiO2 ( x − 1)H 2 O + Heat Tricalcium s ilicate

Hydrolgel

Tricalcium Aluminate (C3A) Due to fast hydration, this compound forms hydrated tricalcium aluminate and is responsible for the first few days of strength. This reaction is highly exothermic, hence cement is made with less C3A. At that time, it generates less heat, develops higher strengths and shows greater resistance to sulphate attacks. 3CaO ⋅ Al 2 O3 + 6H 2 O → 3CaO ⋅ Al 2 O3 + 6H 2 O + Heat Tricalcium aluminate

Hydrated tricalcium aluminate

Due to high heat generation and reactiveness with soil, C3A is the least preferred component in the cement. Tetracalcium Aluminferrite (C4AF) It hydrates very rapidly, reduces clinkering temperature and gives little strength to concrete. 4CaO ⋅ Al 2 O3 ⋅ Fe2 O3 + 7H 2 O → 3CaO ⋅ Al 2 O3 + 6H 2 O + CaO ⋅ Fe2 O3 ⋅ H 2 O

Tetracalcium aluminate

Calcium aluminate

Calcium ferrite

Importance of Gypsum in Cement Tricalcium aluminate (C3A) combines with water very rapidly with the evolution of large amount of heat (exothermic reaction). C3 A + 6H 2 O → C3 A ⋅ 6H 2 O + Heat After the initial set, the paste becomes almost stiff. However, in the presence of gypsum, it reacts with tricalcium aluminate to form insoluble tricalcium sulphoaluminate which helps to retard the speed of the initial set and does not show any tendency to rapid hydration. 3CaO ⋅ Al 2 O3 → 32CaSO 4 ⋅ 2H 2 O → 3CaO ⋅ Al 2 O3 ⋅ 32CaSO 4 ⋅ 2H 2 O Tricalcium aluminate

Gypsum

Tricalcium sulphoaluminate

The aforementioned reaction prevents high concentration of alumina in the cement solution and retards the early initial set of the cement.

7.3.5 Setting and Hardening of Cement When water is added and mixed to cement to form cement paste, hydration begins and it is converted into gel and crystalline material. Solidification, interlocking and binding of the aggregates into a rocklike matter is two-step process in the form of setting and hardening.

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7.14 Engineering Chemistry Setting It is the stiffening of the cement paste with the formation of gel setting that is divided into initial setting and final setting. Initial setting: It refers to the hydration and gel formation from the different constituents of cement. C3 A + 6H 2 O → C3 A ⋅ 6H 2 O + Heat

Tricalcium aluminate

Crystalline hydrated tricalcium aluminate

C4 AF + 7H 2 O → C3 A ⋅ 6H 2 O + CF ⋅ H 2 O + Heat Tetracalcium alumine ferrite

Crystalline

gel

During the hydration of dicalcium, silicate gives tobermorite gel, which possesses very high surface area and a high adhesive property. 2C2S+ 4H 2 O → C3S2 ⋅ 6H 2 O + Ca(OH)2 + Heat

Dicalcium silicate

Tobermonite gel

Crystalline

Final setting: It is the complete formation of tobermorite gel. Hardening Hardening is the development of strength with the crystallisation of calcium hydroxide and hydrated tricalcium aluminate. 2C3S+ 6H 2 O → C3S2 ⋅ 3H 2 O + 3 Ca(OH)2 + Heat

Tricalcium silicate

Tobermonite gel

Crystalline calcium hydroxide

3C3 A + 6H 2 O → C3 A ⋅ 6H 2 O + Heat

Tricalcium aluminate

Crystalline hydrated tricalcium aluminate

Two theories are proposed for explaining the hardening of the cement. Colloidal Theory by Michaels According to this theory, silicate gels are formed with hydration and are responsible for hardening. Crystalline Theory by Le Chatelier According to this theory, crystalline products are formed with hydration, undergo interlocking and are responsible for hardening.

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Hence, setting and hardening of cement is due to the interlocking crystallisation of gels, which are formed by hydrolysis of constitutional ingredients (Figure 7.8). Unhydrated cement + Water Hydration Plastic mass (cement paste) Colloidal theory Metastable gel

Setting

Hardening

Crystaline theory Hydrated crystaline products

Crystaline products Stable gel (Coarser dimensions)

Figure 7.8 Schematic diagram of setting and hardening of cement Ingredients and Reaction Sequence During Setting and Hardening of Cement When water is added to cement, various ingredients undergo hydration and crystallisation in different rates (Figure 7.9). Cement 1 day + water

Hydration Gelation of of C3A 7 days Gelation 28 days C2S S of C and 3 and C3S C4AF

Figure 7.9 Sequence of changes during setting and hardening of cement

7.3.6 ISI Specifications of Cement According to ISI 269–1975, the composition of ordinary Portland cement shall satisfy the following conditions: Chemical Requirement of Cement (i) Ratio of the percentage of lime (CaO) to that of silica (SiO2), alumina (Al2O3) and iron oxide (Fe2O3) when calculated by the following formula: CaO − 0.7SO3 2.8SiO 2 +1.2Al 2 O3 + 0.65Fe2 O3

shall not be less than 1.02.

(ii) Ratio of percentage of alumina (Al2O3) to that of iron oxide (Fe2O3) shall not be less than 0.66. (iii) Weight of insoluble residue shall not exceed 2%. (iv) Weight of magnesia shall not be more than 6%.

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7.16 Engineering Chemistry (v) Total sulphur contents, calculated as sulphuric anhydride (SO3) shall not be more than 2.75%. (vi) Total loss on ignition shall not exceed 4%. Physical Requirements of Cement (i) Setting time: Initial: not less than 30 minutes; final: not more than 600 minutes. (ii) Compressive strength (of 1:3 cement mortar cube of cement ampers; Ennore sand): Three days—not less than 1.6 kg/mm2 (or 16 N/mm 2) Seven days—not less than 2.2 kg/mm 2 (or 22 N/mm 2) (iii) Soundness: By Le Chatelier’s method, it expresses the expansivity of the cement set in 24 hours between 25°C and 100°C. Unaerated cement—maximum 10 mm Aerated cement—maximum 5 mm (iv) Fineness: Not less than 215 m 2/kg; finer the grinding, the greater is the rate of reactions, thereby hastening the early development of strength. However, finer cement generates heat quickly, thereby the cement mortar/concrete is likely to develop cracks.

7.3.7 Analysis of Cement The quality of cement is maintained by conducting various tests from the raw material stage right up to the packing stage, at every half an hour to one hour intervals. In fact, various physical and chemical characteristics are tested. The quality of a sample of cement is determined from a number of measurements as follows. Soundness: The soundness of cement can analysed by Le Chatelier technique and autoclave method. According to the ISI specifications, following properties should attain good quality of cement. Soundness: By auto clave method: Expansion not more than 0.8% By Le Chatelier method: For Aerated cement: maximum 5 mm For Unaerated cement: maximum 100 mm Fitness: By Turbidmetic method: 1600 cm2/gm. By plain permeability method (as specific surface) ≥215 m2/kg. Compressive strength: As per ISI specifications. Tensile strength: As per ISI specifications. Specific gravity: Specific gravity should be 3.1–3.2.

7.3.8 Plaster of Paris/Gypsum Plaster Gypsum (hydrated calcium sulphateCaSO 4 ⋅ 2H 2 O) is extensively used as a raw material for the manufacture of plates, which are almost universally used for coating the inner walls of dwellings. 1 It is the hemihydrate of calcium sulphate 2CaSO 4 ⋅ H 2 O(or CaSO 4 ⋅ H 2 O) 2

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Preparation It is produced by heating pure gypsum to a temperature of about 120°C–160°C. If gypsum is heated above 200°C, anhydrous sulphate is produced, which loses the power of readily combining with water. 1 °C CaSO 4 ⋅ 2H 2 O CaSO 4 ⋅ 2H 2 O CaSO 4 ⋅ H 2 O 200  → CaSO 4 2 Gypsum

Orthorhombic dihydrate

Plaster of paris

The preparation of plaster of Paris from gypsum consists of the following operations: (i) Crushing and grinding of gypsum (ii) Calcination of ground gypsum in kilns by heating about 150°C (iii) Pulverising the calcined product Setting and Hardening Plaster of Paris forms a plastic mass when it is mixed with water. This plastic mass quickly sets or hardens, expanding in the process and regains the closely packed crystalline structure of gypsum. The setting of plaster of Paris can be accelerated by mixing it with alkali sulphates such as K 2SO4, Na2SO4 or alums, which initiate as well as hasten the crystallisation process. Applications (i) Its slight expansion on setting renders plaster of Paris suitable for making mould. Therefore, some details are accurately reproduced. (ii) It is used in making surgical bandages, structural tiles and castings. (iii) It is used as plaster for walls and in plaster boards, made up of alternate layers of a fibrous material such as felt or paper.

7.4 REFRACTORIES Refractories are ceramic materials that can withstand high temperatures as well as abrasive and corrosive actions of molten metals, slags and gases, without suffering a deformation in shape. The main objective of a refractory is to confine heat.

7.4.1 Characteristics of Good Refractory Materials (i) A good refractory material should have excellent heat, corrosion and abrasion resistance. (ii) It should possess low thermal coefficient of expansion and should expand and contract uniformly with increase and decrease of temperature, respectively. (iii) It should possess high fusion temperature. It should be infusible at operating temperatures. (iv) It should be able to withstand the overlying load of structure, at operating temperatures. (v) It should be chemically inert towards corrosive action of molten metal, gases and slags produced in its immediate contact with furnaces. (vi) It should not crack at operating temperatures.

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7.4.2 Failures of Refractory Materials If a given refractory material does not have the aforementioned characteristics, it will fail in service. Thus, we can easily summarise conditions, which lead to failure of refractory materials as follows: (i) Refractory material which does not have resistance to required heat, corrosion and abrasion. (ii) Refractory material which has higher thermal expansion. (iii) Refractory material which has less refractoriness than the operating temperature. (iv) Lower quality refractory bricks than the actual load of raw materials in products. (v) Usage of basic refractory material in a furnace in which acidic reactants and/or products are being processed and vice versa (vi) Refractory material that undergoes considerable volume changes during their use at high temperatures.

7.4.3 Classification of Refractories On the basis of chemical properties, refractories are broadly classified into three main categories. Acidic Refractories Refractories which consist of acidic materials are known as acidic refractories. They are easily attacked by basic materials and not by acidic materials. For example, alumina, silica and fireclay refractories. Basic Refractories Refractories which consist of basic materials are known as basic refractories.There are easily attacked by acidic materials and not by basic materials. For example, magnesite and dolomite. Neutral Refractories Refractories which consist of weak acidic/basic materials are known as neutral refractories. For example, zirconia, graphite, chromite and carborundum.

7.4.4 Properties of Refractories The important properties of refractories are as follows. Refractoriness It is the ability of a material to withstand heat without appreciable deformation. It is commonly measured as the softening or melting temperature of the material. The softening temperatures of refractory materials are determined by using “pyrometric cones (seger cones) test” (Figure 7.10). The refractory should have a softening temperature much higher than the operating temperature of the furnace in which it is to be used.

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Refractoriness is generally determined by comparing the behaviour of heat on the cone of the material to be tested with that of a series of seger cones of standard dimensions. Refractoriness is expressed in terms of pyrometric cone equivalent. Cones are 38mm height, 19mm long sides with triangular base pyramids, and at definite temperatures, they can melt or fuse. The temperature at the apex touching the base is indication of fusion/softening of the test cone. The number of the standard cones fusing along with the test cone is the pyrometric cone equivalent (PCE) of that particular refractory. If the test cone fuses later than one standard cone and earlier than the next cone, the PCE is the average value of the two.

Softened and deformed apex joint touching the base Plaque

Figure 7.10 Seger cone test Porosity Porosity is the property of a solid which contains openings, spaces or minute channels. It can be expressed as follows: P=

W −D ×100 W−A

Where W is the weight of saturated specimen D is the weight of dry specimen A is the weight of saturated specimen submerged in water If the refractory has pores, the entry of gases, slags, etc., is easy and can react up to a greater depth. This can reduce the life of the refractory material. Consequently, it can affect many important properties of the refractory such as decreasing the strength, resistance to corrosion, resistance to abrasion but increased resistance to thermal spalling. Hence, a good refractory should have low porosity. Strength or Refractoriness–under Load The refractory material must possess high mechanical strength, even at operating temperatures to bear the maximum possible load, without breaking. Dimension Stability It is the resistance of a material to any volume changes, which may occur on its exposure to high temperature, over prolonged time. It may reversible or irreversible. Chemical Inertness A refractory does not easily form fusible products with gases, ash, slags, etc., and hence should be chemically inert.

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7.20 Engineering Chemistry Thermal Expansion A refractory material should have the least possible thermal expansion due to the following reasons: (i) Expansion of a refractory decreases the capacity of the furnace. (ii) Repeated expansion and contraction contribute much towards rapid breakdown and wear and tear of the material structure. Thermal Conductivity Depending upon the type of furnace refractory, materials of high and low thermal conductivity are required. In most cases, the furnace is lined with refractories of low-heat conductivities to reduce heat losses externally by radiation; otherwise, maintenance of high temperatures inside the furnace will become difficult. In muffle furnace walls and coke-oven batteries, a good heat conductivity of refractory is desirable for effective heat transmission. Thermal Spalling The breaking, cracking, peeling off or fracturing of a refractory brick or block under high temperature is known as thermal spalling. Thermal spalling may be due to the following: (i) Rapid change in temperature (ii) Slag penetration into the refractory brick Thermal spalling can be decreased by taking the following precautions: (i) (ii) (iii) (iv)

Using high porosity, low coefficient of expansion and good thermal conductivity refractory bricks Avoiding sudden temperature changes By overfiring the refractories By modifying the furnace design

Heat Capacity It depends on the following: (i) Thermal conductivity (ii) Specific heat and (iii) Specific gravity of refractory Resistance to Abrasion or Corrosion Refractoriness is desirable that least abraded by descending hard charge, flue gases escaping at high speed, particles of carbon or grit, etc. Electrical Conductivity Refractories specially used for lining electric furnaces should have low electrical conductivity. Except graphite, all refractories are poor conductors.

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Permeability The rate of diffusion of gases, liquids and molten solids through a refractory is known as permeability. It mainly depends on the size and number of connected pores. Permeability increases with temperature. Texture Due to large porosity, coarse- or light-textured bricks are less in weight; hence, they are more resistance to sudden temperature changes.

7.4.5 Manufacture of High-Alumina Bricks, Magnesite Bricks and Zirconia Bricks High-alumina Bricks High-alumina bricks are made by mixing calcined bauxite (Al2O3) with clay bind, which contains 50% or more of Al2O3. Properties High-alumina bricks have very low coefficient of expansion, high porosity, great resistance to slags, very little tendency to spall high temperature, load-bearing capacity, excellent wear-resistance and stability, both in oxidizing and reducing conditions and are particularly inert to the action of gasses such as CO2, H2 and natural gas. Thus, they are very good refractories, but very expensive; hence, their use is limited. Uses Medium-duty Bricks Bricks which contain 50–60% Al2O3 are used for zones of vertical shaft kilns for burning limes, linings of Portland cement rotary kilns, soaking pits, reheating furnaces, hearths and walls, etc., and are subject to high abrasion. High-duty Bricks Bricks which contain 75% Al2O3 are used in sintering or the hottest zones of cement rotary kilns, lower parts of soaking pits, brass melting reverberatories, lead dressing reverberatory furnaces, aluminium melting furnaces, combustion zones of oil-fired furnaces, etc. Magnesite Bricks Magnesite bricks are the most widely used basic refractories. Calcined magnesite is powdered to a proper size, mixed with binding material as caustic magnesia or iron oxide and the mixture is grounded with water moulded into bricks, then slowly heated to 1,500°C upto eight hours and then slowly cooled. Properties Magnesite bricks can be used without load up to 3,000°C and with load of 3.5 kg/cm2 up to 1,500°C. They possess good crushing strength, good resistance to basic slag and less shrinkage. They have a lot

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7.22 Engineering Chemistry of spalling with sudden temperature changes and their resistance to abrasion is poor. They have lot of tendency to combine with water and CO2. Uses These are mainly used in open-hearth furnaces where high temperature required. They are also used in hot mixer linings, copper converters, reverberatory furnaces for smelting antimony, copper, lead, etc., ores, refining furnaces and hot zones of cement rotary kilns. Zirconia Bricks These are prepared by heating zirconite mineral (ZrO2) and colloidal zirconite or alumina as binding material at 1,700°C. This is stabilised by adding of MgO or CaO without undergoing any volume changes on heating and cooling. Properties Zirconia bricks are usually known as neutral refractories, but they have no resistance to acids, slags, etc. Hence, they are between neutral and basic refractories. Without load, they can withstand up to 2,000°C, but specially prepared bricks can be used up to 2,600°C and with load of 3.5 kg/cm2, up to 1,900°C. They have good resistance to thermal shocks. Uses As these bricks are very costly, they are used only in high frequency electric furnaces.

7.5 LUBRICANTS In all machines, lot of wear and tear is observed due to friction. Therefore, a large amount of energy is also lost in the form of heat and moving parts get heated and damaged. The ill-effects of frictional resistance can be minimised by using a suitable substance called lubricant, which can form a thin layer in between the moving parts and keep the sliding or moving surfaces apart. Hence, frictional resistance and consequent destruction of material is minimised. “The process of minimising frictional resistance between moving or sliding surfaces by the introduction of lubricants in between them is called lubrication”.

7.5.1 Important Functions of Lubricants (i) It avoids direct contact between the rubbing surfaces and reduces surface deformation, wear, tear and seizure. (ii) It acts as a coolant by reducing loss of energy in the form of heat. (iii) It enhances efficiency of a machine by reducing wastage of energy and expansion of metal by local frictional heat. (iv) It avoids seizure and relative motion of moving surfaces, such that running cost of the machine will be reduced.

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(v) The lubricant used between piston and the cylinder wall of an internal combustion engine acts as a seal and can prevent the leakage of gases from the cylinder under high pressure.

7.5.2 Mechanism of Lubrication There are mainly three types of mechanism by which lubrication is done. These are explained here. Thick Film, Fluid Film or Hydrodynamic Lubrication In this mechanism, moving or sliding surfaces are separated by thick film of lubricant fluid; hence, it is known as thick film or fluid film lubrication. The thick film of lubricant covers entire moving surfaces and fills irregularities. Load Thick layer of Therefore, there is no direct contact between the surlubricant faces of machine and consequently it reduces the wear. This Velocity is shown in Figure 7.11. Here, only the internal resistance is observed between the particles of lubricant; hence, the chosen lubricant should have minimum viscosity under the working conditions. Hydrodynamic friction occurs in the case of shaft running places like journal bearings, which is shown in Figure 7.12. Thick film lubrication hydrocarbon oils are considered satisfactory lubricants. Hydrocarbon lubricants are blended with selected long-chain polymers to maintain viscosity of the Figure 7.11 Fluid-film lubrication oil throughout the year. Bearing lining

Boundary or Thin Film Lubrication In this kind of lubrication, moving surfaces are separated by a thin layer of lubricant, which is absorbed by physical or chemical forces on the metallic surfaces as shown in Figure 7.13. Here, the continuous film of lubricant cannot persist due to any of the following reasons: (i) (ii) (iii) (iv)

Rotating staff in floating lubricant Thick film of lubricant

Figure 7.12 Hydrodynamic lubrications

A shaft starts moving from rest The speed is very low The load is very high The viscosity of the oil is very low

Vegetable oils, animal oils and their soaps possess the property of adsorption either physically or chemically to the metal surfaces and form a thin film of metallic soap, which acts as a good lubricant. Fatty oils possess greater adhesion property than mineral oil, and to improve the oiliness of mineral oils, a small amount of fatty oils is added. Graphite and molybdenum disulphide are also used for boundary lubrication.

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Load Velocity

Adsorbed thin layers of lubricant (vegetable or animal oil)

Figure 7.13 Boundary lubrication

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7.24 Engineering Chemistry Extreme Pressure Lubrication In this mechanism, the moving or sliding surfaces are under very high pressure and speed; hence, this is known as extreme pressure lubrication. Under such conditions, a high local temperature is attained, and liquid lubricants fail to stick and may decompose or vaporise. Special additives are added to mineral oils to meet the extreme pressure conditions and are called extreme pressure additives. Organic compounds having active radicals or groups such as chlorine, sulphur or phosphorous act as good additives. These compounds react with metallic surfaces to form metallic chlorides, sulphides or phosphides as more durable films, capable of withstanding very high loads and temperatures.

7.5.3 Classification of Lubricants On the basis of their physical state, lubricants can be classified into three categories as listed hereunder. Liquid Lubricants or Lubricating Oils Apart from reducing friction and wear, lubricating oil also acts as a cooling medium sealing agent, corrosion preventer, etc. According to origin, lubricating oils are classified into animal and vegetable oils, mineral or petroleum oils and blended oils. Animal and Vegetable Oils Vegetable and animal oils possess good oiliness but they are costly, undergo oxidation easily, forming gummy and acidic products, get thickened on coming in contact with air, etc. Hence, they are rarely used as lubricant, but are used as blending agents. Mineral or Petroleum Oils Mineral oils are mainly obtained by the distillation of petroleum. These are widely used lubricants because they are cheap, abundantly available and quiet stable under service conditions. The hydrocarbon oil chain length varies between 12 and 50 carbon atoms. The shorter chain hydrocarbons have lower viscosity than longer chain hydrocarbons. When compared to animal and vegetable oils, mineral oils possess poor oiliness; therefore, to increase oiliness, high molecular weight compounds such as oleic and steric acids are added. Blended Oils In many modern machinery, no single oil serves as the most satisfactory lubricant. Improving important properties by incorporating specific additives is known as blending of oils; such oils give the desired lubricating properties. Properties of a Good Lubricating Oil A good lubricating oil must possess the following qualities: (i) Low pressure (ii) High boiling point

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(iii) Adequate viscosity to particular service conditions (iv) Low freezing point (v) High oxidation resistance (vi) Heat stability (vii) Non-corrosive property and (viii) Stability to decomposition at the operating temperatures.

Greases or Semi-solid Lubricants Semi-solids consisting of soap dispersed throughout liquid lubricating oil is grease; the liquid lubricant may be a petroleum oil or even a synthetic oil, and may contain any of the additives for specific requirements. Preparation Greases are prepared by the saponification of fat with alkali, followed by adding hot lubricating oil under agitation. The total amount of mineral oil added determines the consistency of the finished grease. The structure of lubricating greases is like that of a gel. Soaps are gelating agents, which give an interconnected structure by intermolecular forces containing the added oil. The soap dissolves in the oil at high temperature; hence, inorganic solid, thickening agents are added to improve the heat resistance of grease. Greases have higher shear or frictional resistance than oils and can support much heavier loads at lower speeds. Greases are used in the following situations: (i) Where oil cannot remain in place due to high load, low speed, intermittent operation, sudden jerks, etc., for example in-rail axle boxes. (ii) Bearing and gears that work at high temperatures. (iii) Where the bearing needs to be sealed against entry of dust, dirt, grit or moisture because greases are less liable to contamination by these. (iv) Where dripping or spurting of oil is undesirable, because unlike oils, greases do not splash or drip over articles being prepared by the machine. For example, paper manufacturing machines, textiles, edible articles, etc. The main function of a soap is to acts as a thickening agent; hence, grease sticks firmly to metal surfaces. The nature of the soap decides its consistency, resistance to water and oxidation and temperature up to the grease can be used. Hence greases are classified according to usage soap in their manufacture. Some of the important greases are as follows: Calcium-based Greases or Cup-greases Calcium-based greases are emulsions of petroleum oils with calcium soaps, generally, prepared by adding required amount of calcium hydroxide to hot oil while under agitation. These are the cheapest and the most commonly used greases. They are insoluble in water and are water-resistant. They are satisfactory for use at low temperatures, because above 80°C, oil and soap begin to separate out.

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7.26 Engineering Chemistry Soda-base Greases These are petroleum oils, thickened by mixing sodium soaps. They are not water-resistant, because the sodium soap content is soluble in water. They can be used up to 175°C and are suitable for use in ball bearings, where the lubricant gets heated due to friction. Lithium-based Greases They are emulsions of petroleum oils with lithium soaps. They have high water resistance and are suitable only below 15°C. Axle Greases They are cheap resin greases and are prepared by adding lime or heavy metal hydroxide to resin and fatty oils. The resulting mixture is thoroughly mixed, allowed to stand and tack or mica-like fillers are finally added. These are water-resistant and are also suitable for less delicate equipment working under heavy loads at low speed. Besides these, there are greases prepared by dispersing solids (like graphite, soapstone, etc.,) in mineral oil. Solid Lubricants Graphite and molybdenum disulphide are the important solid lubricants. These are used in the following conditions. (i) The operating temperature or load is too high. (ii) The blended lubricating oil or the mixed grease is unacceptable. (iiii) There is a need to avoid combustible lubricants. The layered structure of graphite and the sandwich-like structure of molybdenum disulphide are shown in Figure 7.14 (a) and (b). Each carbon atom is bonded by only three covalent bonds

Carbon atoms in a network of hexagons

3.08 Å Mo atom layer

Plates of layers capable of sliding parrellel to each other 3.4 Å

3.13 Å S atom layer 3.13 Å Mo atom layer

1.42 Å (a)

(b)

Figure 7.14 (a) Layered structure of graphite (b) Sandwich-like structure of molybdenum disulphide

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Hence, the force to shear the crystals parallel to the layers is low and consequently, the parallel layers slide over one another easily. Usually, some organic substances are mixed with solid lubricants so that they may stick firmly to the metal surface. Solid lubricants are used either in the dry powder form or mixed with water or oil. Graphite is the most widely used lubricant because it is very soapy to touch, non-inflammable and not oxidised in air below 375°C. Graphite is used in the form of powder or suspension in oil or water with the help of emulsifying agent tannin. Graphite is dispersed in oil is called oildag and when dispersed in water, it is called “aquadag”. In the absence of air, it can be used up to very higher temperature. Graphite is used either in powdered form or as suspension. Graphite greases are used at higher temperature.

7.5.4 Properties of Lubricants The properties of lubricants are described here. Neutralisation Number The acidity or alkalinity of a lubricating oil is determined in terms of its neutralisation number. Determination of acidity is more common and is expressed as the acid value or acid number. It is defined as the “number of milligrams of potassium hydroxide required to neutralise all the free acid present in one gram of the lubricating oil.” Even the most carefully refined oil may have slight acidity. This is due to the presence of minute amount of organic constituents that are not completely neutralised during the refining treatment or due to traces of residues from the refining process. This small intrinsic acidity may not be harmful in itself, but the degree to which it increases in already used oil is usually taken as a measure of the deterioration of the oil due to oxidation or contamination. In fact, acid number greater than 0.1 is usually taken as an indication of oxidation of the oil. Saponification Number The saponification value of an oil is defined as the “number of milligrams of potassium hydroxide required to saponify one gram of the oil”. This is usually determined by refluxing a known quantity of the oil with a known excess of standard KOH solution and determining the alkali consumed by titrating the unreacted alkali. Animal and vegetable oils undergo saponification but mineral oils do not. Further most of the animal and vegetable oils process their own characteristic saponification values. Hence, the determination of the saponification value helps in ascertaining the presence of animal and vegetable oils in a lubricant. Conversely, since each of the fined oil has its own specific saponification number, any deviation from this value in a given sample indicates the probability and extent of adulteration. Aniline Point The aniline point of an oil is defined as “the minimum equilibrium solution temperature for equal volumes of aniline and oil sample”. Aromatic hydrocarbons have high tendency to dissolve natural and synthetic rubbers; this tendency can be determined on the basis of aniline point of an oil. A higher aniline point means lower percentage of hydrocarbons; therefore, having higher aniline point is desirable.

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7.28 Engineering Chemistry The aniline point is determined by thoroughly mixing equal volumes of aniline and the oil sample in a tube and heating the mixture until a homogeneous solution is obtained. This is allowed to cool at a specified rate until the two phases just separate out. The temperature corresponding to this particular observation is reported as the aniline point. Cloud Point and Pour Point Petroleum oils are complex mixtures of chemical compounds and do not show a fixed freezing point. When they are sufficiently cooled, they become plastic solids due to the formation of solid crystals or the congealing of the hydrocarbons present. “The cloud point is the temperature at which this crystallisation of solids in the form of a cloud or haze first becomes noticeable,” when the oil is cooled in a standard apparatus at a standard rate. The pour point is “the temperature at which the oil just ceases to flow when cooled at a standard rate in a standard apparatus”. The pour point has a greater significance for lubricating oil because it determines the suitability of a hydraulic oil for low temperature installations. Refrigerator plants, air-craft engines, etc., are some important examples, which may be required to start and operate at sub-zero temperatures. Flash Point and Fire Point The flash point of an oil is defined as “the minimum temperature at which the oil gives off sufficient vapour to ignite momentarily when a flame of standard dimensions” is brought near the surface of the oil. The fire point of an oil is defined as “the lowest temperature at which the vapours of the oil burn continuously for at least five seconds” when the standard flame is brought near the surface of the oil. The lubricating oil should have flash point that is reasonably above its working temperature. This ensures safety against fire hazards during usage, storage and transport. The flash point of lubricating oil can be determined by Pensky Marten’s apparatus. Viscosity and Viscosity Index Viscosity is one of the most important properties of lubricating oil. The formation of a fluid film of a lubricant between the friction surfaces and the generation of frictional heat under particular conditions of load, bearing spread and lubricant supply mostly depend upon the viscosity of the lubricant and, to some extent, on its oiliness. When large working clearances exist between the friction surfaces, a high viscosity oil is generally recommended to “cushion” the intermediate application of load. However, it is often necessary to sacrifice some of the cushioning effect of viscous oil by the partial substitution of a thinner oil to provide good circulation of oil to dissipate the frictional heat. If the viscosity of the oil is very low, the fluid lubricant film cannot be maintained between the moving surfaces as excessive wear may take place. On the other hand, if the viscosity of lubricating oil is very high, excessive friction would occur due to shearing of oil. Hence, in hydrodynamic lubrication, the lubricant selected must possess sufficiently high viscosity due to adherence to the bearing and prevent it being squeezed out due to high pressure and yet fluid enough so that the resistance to the shear is not too high. It is, therefore, essential to have knowledge of the viscosity of lubricating oil. Viscosity is a measure of the internal resistance to the motion of a fluid and is mainly due to the forces of cohesion between the fluid molecules. Absolute viscosity can be defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel planes in the fluid unit distance apart. The units of absolute viscosity h (eta) in the centimetre-gram-second (CGS)

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system are poise and centipoise (1/100th of a poise). A poise is equal to one dyne per second per square centimetre. The viscosity of water at 20°C is about a centipoise. The ratio of absolute viscosity to density for any fluid is known as the absolute kinematic viscosity. It is denoted by h, and in the CGS system, its units are stokes and centistokes (1/100th of a stoke). n=

h r

where n = absolute kinematic viscosity h = absolute dynamic viscosity r = density of the fluid. The dimensions of dynamic viscosity are HL −1T−1, and the dimensions of kinematic viscosity are L2T−1. For academic purposes, viscosity is usually expressed in centistoke, but a more common practical measure of the viscosity of an oil is the time in seconds for a given quantity of the oil to flow through a standard orifice under specified set of conditions. Viscosities are usually determined with Redwood viscometer in commonwealth countries, with Engler’s viscometer in Europe and with Saybolt’s viscometer in the USA. In these commercial viscometers, a fixed volume of the liquid is allowed to flow through the standard orifice of particular standard apparatus. Redwood (No. 1) seconds at 25°C. The viscosity of the oil so determined in the time unit is sometimes called relative viscosity. Since the instruments used are of standard dimensions, the kinematic viscosity of the oil in centistokes can be calculated from the time taken by the oil to flow through the standard orifice of the instrument, with the help of the following equations: m = Ct (for fluids whose kinematic viscosity is more than centistokes) and m = Ct − b /t (for fluids having kinematic viscosities lesser than or equal to 10 centistokes) m = Kinematic viscosity in centistokes t = Time of flow in seconds C = Viscometer constant B = Coefficient of kinetic energy, which may be determined experimentally or eliminated by choosing long flow times For routine purposes, the test viscometer may be calibrated and the constant C determined by using solutions of known viscosity. The primary standard used is freshly distilled water whose kinematic viscosity is 1.0008 centistokes. Other standards usually employed are as follows: 40% sucrose solution: n = 4.390 cs at 25°C, r = 1.17395 60% sucrose solution: n = 33.66 cs at 25°C, r = 1.28335 For Redwood No. 1 viscometer, the values for the constants are as follows: Time of flow, t 40–85 seconds 85–2,000 seconds

B 190 65

C 0.264 0.247

These constants are based on the results of the work carried out at the National Physical Laboratory at a temperature of 70°F (21.11°C) and with the ranges of viscosity; at that temperature, the results are accurate to ±1%.

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7.30 Engineering Chemistry Redwood No. 2 viscometer is used for every viscous liquids and gives one-tenth the value of Redwood No. 1 viscometer. Viscosity Index The viscosity of an oil decreases with increase in temperature as a result of decrease in intermolecular attraction due to expansion. Hence, it is always necessary to state the temperature at which the viscosity is determined. In many applications, the lubricating oil will have to function in a machinery over considerably wide range of operating temperatures. If this occurs due to seasonal variations in atmospheric temperature, adjustments can be affected by selecting different oils of appropriate viscosity for different seasons. However, in case of internal combustion engines, aeroplanes, etc., the lubricant used must function at low starting temperature as well as at very high operating temperature. Since the viscosity of lubricating oils decreases with temperature, it is impossible to select an oil having same viscosity over such a wide range of operating temperatures. However, one can select an oil whose variation in viscosity with temperature is minimum. This variation can be indicated either by viscosity temperature curves or by means of the viscosity index. The viscosity index is the numerical expression of the average slope of the viscosity temperature curve of lubricating oil between 100°F to 210°F. The oil under examination is compared with two standard oils having the same viscosity at 210°F as the oil under test. Oils of the Pennsylvanian type crudes thin down the least with an increase in temperature, whereas oils of the Gulf coast origin thin down the most as the temperature is increased. Hence, the viscosity index of Pennsylvanian oil is taken as 100 and that of the Gulf oil as zero. Then, the viscosity of the oil under investigation is deducted as follows: V − VX Viscosity index of the oil under test L ×100 VL − VH where V L = viscosity at 100°F of Gulf oil standard, which has the same viscosity at 210°F as that of oil under test VX = viscosity of the oil under test VH = viscosity at 100°F of Pennsylvanian standard oil, which has the same viscosity at 210°F as that of oil under test. Thus, the higher the viscosity index, the lower the rate at which its viscosity decreases with increase in temperature. Hence, oils of high viscosity index, that is, those having that viscosity temperature curves are demanded for air-cooled internal combustion engines and aircraft engines. In general, oils of high specific gravity have steeped viscosity temperature curves. However, all oils tend to attain the same viscosity above 300°C. By and large, light oils of low viscosity are used in plain bearings for high-speed equipment such as turbines, spindles and centrifuges, whereas high viscosity oils are used with plain bearings of lowspeed equipment. Mechanical Stability Four balls extreme-pressure test is one of the important mechanical tests to judge the suitability of a lubricant under conditions of very high pressure, as shown in Figure 7.15. In this test, the lubricant

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under test is powered in a machine containing four balls. Here, the upper ball is rotated and the lower three balls are stationary. The load is gradually increased, the ball is withdrawn and examined at specific intervals for scale formation, etc., and under a given load, the ball bearings after the test comes out clean if the lubricant desirable. When the load is progressively increased, the liberated heat welds the ball together. Here, the lubricant is said to have completely failed. Hence, this test enables us to determine the maximum load that can be carried safely by a lubricant. The four-balls extreme-pressure lubricant tester is shown in Figure 7.15.

Top ball rotates Stationary balls Lubricant sample under-test Load force

Figure 7.15 Four-balls extreme-pressure lubricant tester

7.5.5 Redwood Viscometer The Redwood viscometer is made in two sizes. The Redwood No. 1 viscometer is commonly used for determining viscosities of lubricating oils and has an efflux time of 2,000 seconds or less. The Redwood No. 2 viscometer is similar to the Redwood No. 1 type but the jet for the outflow of the oils is of a larger diameter and hence gives an efflux time of approximately one-tenth of that obtained with Redwood No. 1 instrument under otherwise identical experimental conditions. The Redwood No. 2 instrument is therefore used for the oils having higher viscosities such as fuel oils. The Redwood viscometer does not give a direct measure of viscosity in absolute units but it enables the viscosities of oils to be compared by measuring the time of efflux of 50 ml of oil through the standard orifice of the instrument under standard conditions. The results given by these two viscometers are reported as Redwood No. 1 viscosity or Redwood No. 2 viscosity followed by the efflux time in seconds of the experimental temperature. Description The Redwood No. 1 viscometer as shown in Figure 7.16 essentially consists of a standard cylindrical oil cup made up of brass and silvered from inside and has 90 mm height and 46.5 mm in diameter. The cup is open at the upper end. It is fixed with an agate jet in the base. The diameter of the orifice is 1.62 mm and the internal length is 10 mm. The upper surface of the agate is ground to concave depression into which a small silver-plated brass ball is attached to a stout wire can be placed in such a way that the channel is totally closed and no leakage of the oil from the cup through the orifice

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7.32 Engineering Chemistry Thermometers

Stirrer blade Metallic oil cup

Pointer

Water path

Stirrer shield Valve rod

Oil

Heating tube Water

Kohilrausch flask 50 mL

Water

Agate jet of 1.82 mm dia and 10 mm length

Levelling screw

Figure 7.16 Redwood viscometer No. 1 can take place. The cup is provided with a pointer, which indicates the level up to which the oil should be filled in a cup. The lid of the cup is provided with an arrangement to fix a thermometer to indicate the oil temperature. The oil cup is surrounded by a cylindrical, copper vessel containing water, which serves as a water bath used for maintaining the desired oil temperature with the help of electrical heating oils or by means of a gas burner as the case may be. A thermometer is provided to measure the temperature of water. A stirrer with four blades is provided in a water bath to maintain uniform temperature in the bath, thus enabling uniform heating of the oil. The stirrer contains a broad, curved flange at the top to act as a shield for preventing any water from splashing into the oil cylinder. The entire apparatus rests on a sort of tripod stand provided with levelling screws at the three legs. The water bath is provided with an outlet for removing water as and when needed. A sprit level is used for levelling the apparatus and a 50 ml flask for receiving the oil from the jet outlet is also provided. Working The instrument is levelled with the help of the levelling screws on the tripod. The water bath is filled with water to the height corresponding to the tip of the indicator up to which the oil is to be filled in the cylindrical cup. The orifice is sealed by keeping the brass ball in position. Then the oil under test is carefully poured into the oil cup up to the tip of the indicator. The 50 ml flask is placed in position below the jet. The oil and water are kept well-stirred and the respective temperatures are noted. The ball is raised and suspended from the thermometer bracket. Simultaneously, a stopwatch is started. When the level of the oil dropping into the flask just reaches the 50 ml mark, the stopwatch is stopped and the time is noted in seconds. The ball value is replaced in the original position to prevent the overflow of the oil. The experiment is repeated, and the mean value of time of flow for 50 ml of the oil is reported as t seconds, Redwood 1 at T°C. The usual test temperatures stipulated are 21.11°C (70°F), 60°C (140°F) and 93.33°C (200°F).

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During the test, the measuring flask should be shielded from draughts with the help of metal shields usually supplied with the instrument.

7.5.6 Engler’s Viscometer This instrument is diagrammatically presented in Figure 7.17. The water bath is heated by a gas ring, and its temperature is kept uniform with the help of the stirrer. The oil cylinder is fitted with three gauge points, which indicate the amount of oil required and also serve as a means of levelling the instrument. The loosely fitting cover carrying thermometer can be gently rotated to agitate the oil. The jet is slightly tapered and is made of platinum for standard work and nickel for general work. The valve pin, which seats itself in the jet, is lifted at the commencement of a test and is supported in the cover by a cross pin. As the valve pin is lifted, the stopwatch is started and the time of outflow of 200 ml of the oil is determined. The viscosity is expressed in Engler degrees or degree E by using water as standard. The time of outflow of 200 ml of water at 20°C is taken as 52 seconds. The viscosity in degrees E is calculated by dividing the time m seconds for the outflow of 200 ml of oil by time of outflow of 200 ml of water at 20°C. Bath thermometer

Valve pin

Stirrer

Bath liquid

Oil

200 ml

G-Gauge points (three altogether)

Gas ring

Figure 7.17 Engler’s viscometer

7.5.7 Saybolt Viscometer A single unit Saybolt universal viscometer is shown in Figure 7.18. In a multiple-unit viscometer, a number of oil cups can be accommodated in the same bath, thus enabling tests on a number of oils to proceed at the same time. Instruments can be fitted with an electric immersion heater, a U-tube for steam heating or water cooling and a gas ring, which is placed inside the air jacket surrounding the water bath. The bath liquid is stirred by rotating the cover by means of the two handles as a turn-table arrangement.

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7.34 Engineering Chemistry Bath thermometer Gallery

Electric heater U-Tube

Stirrer blade Oil

Stirrer blade Air jacket Cork

Gas-ring

60 CC

Oil

Stirrer handle Electric heater

Top view

Figure 7.18 Saybolt viscometer The temperature can be regulated by running cold or warm water through the U-tube irrespective of the heating arrangement used. The jet is made of a hard non-corrodible metal such as monel or stainless steel. The lower end of the jet opens into a larger tube. This tube, when stoppered by a cork, becomes a closed air chamber preventing the oil flowing out (Figure 7.18). To start the test, the bath is brought to the test temperature and the oil is heated to the same temperature in a separate vessel. The oil is then poured into the oil cylinder and stirred with the oil thermometer and any excess oil flowing over into the surrounding gallery. When the oil and the bath are at the same temperature, the oil thermometer is removed, the excess oil drawn off from the gallery with a pipette, the cork withdrawn and the stopwatch started. The collecting flask is arranged such that the oil stream will strike its neck, thus avoiding the formation of foam. For very viscous fuels, a viscometer with a larger jet known as the Saybolt furol viscometer is used. The Saybolt universal viscometer can be used for oils having flow times of more than 32 seconds. There is no maximum unit; but in general, for liquids having flow times over 1,000 seconds, the Saybolt furol viscometer is better.

7.5.8 U-Tube Viscometer The standard U-tube viscometer (Figure 7.19(a) and (b)) is an improved form of the Ostwald viscometers, which is used for the determination of the absolute viscosity of lubricating oils. The determination of absolute viscosity of lubricating oils by the U-tube viscometer based on Poiseuille’s law.

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P πr 4 t 8l η

where V = volume of the liquid flowing through a capillary tube of length l (cm) of uniform radius r (cm) in a times t (seconds) and h (poise) is the coefficient of viscosity of the liquid at the particular temperature. Hand pump D

C B

F D

(a)

(b)

Figure 7.19 U-tube viscometers (a) standard U-tube viscometer (b) Ubbelohde suspende level viscometer The determination of absolute viscosity by the U-tube viscometer essentially consists of measurement of the time of passage through the capillary of a fixed volume of liquid under a fixed mean hydrostatic head ρ of the liquid. If the density of the liquid is d, then P ∝ d and since, for a given viscometer, h ∝ td. h = ktd where k is the proportionality constant; It can be determined for different viscometers from its known dimensions or by calibration with a standard liquid such as water or any other liquids. Absolute kinematic viscosity v =

absolute dynamic viscosity ( η) centistokes density of the liquid (ρ)

7.5.9 C onversion of Redwood, Engler and Saybolt Viscosities into Absolute Units Redwood, Engler and Saybolt instruments are not the ideal methods of determining absolute viscosities. The conversion values are only considered good approximations when only taken at the same temperature.

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7.36 Engineering Chemistry Therefore, the conversion of the aforementioned relative viscosities to absolute viscosities is done with the help of the following equation: v = ct − b /t where v = Kinematic viscosity in centistokes, t = time flow in seconds, c and b are constants. Values of c and b are given in Table 7.2. Table 7.2 Values of c and a Instrument

Value of c

Value of b

Redwood No. 1

0.25

172

Redwood No. 2

2.72

1,120

Saybolt universal

0.22

180

Engler

0.147

374

7.6 EXPLOSIVES AND PROPELLANTS Explosives An explosive is a “substance or compound or mixture, which when subjected to thermal and mechanical shock, gets very rapidly oxidised exothermically with a sudden release of potential energy”. The explosive reaction is exothermic, so the products get heated to a high temperature and a high pressure is exerted on the surroundings. The amount of power available from a given weight or volume of explosive, is called “power to weight ratio”.

7.6.1 Some Important Terms about Explosives Some important terms about explosives are as follows: Explosive Strength It is the energy liberated per unit mass of the explosive (cal/g). Velocity of Detonation It is the velocity with which the given explosive detonates. Sensitivity It can be determine the effect and impact of friction, heat, electric spark or detonator wave etc., on explosives. Some explosives may detonate by a feather touch, whereas some may not detonate even with a hammer blow. Sensitivity plays a key role in the selection of explosives for a particular purpose. Brisance It indicates the shattering power of an explosive.

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7.6.2 Classification of Explosives Explosives are broadly classified into three groups. Detonators/Primary Explosives/Initiating Explosives These are highly sensitive, and explode on receiving a slight-shock or when exposed to fire. Hence, they should be handled with utmost care. Some of examples are as follows: Lead Azide (PbN6 ) It is very popular for military uses due to low cost, excellent initiating action and stability in storage. Mercury Fulminate [Hg(CNO2 )] It is more sensitive and expensive than lead azide. It is slightly toxic. Tetracene (C2H7N7O) It requires low initiating action; ignites easily with high heat of explosion and produces a large volume of gases. Diazodinitro Phenol (DDNP) It is quite sensitive and has high brisance and consequently can initiate explosion even in less sensitive high explosives. Propellents or Low Explosives These are simply burns and do not explode all of a sudden. The chemical reactions taking place in such explosives are slow and their burning proceeds from the surface inwards in layers at an approximate rate. The gases evolved disperse readily without building high pressure and consequently, they can be easily controlled. Some of examples are as follows: Gun Powder or Black Powder It is a mixture of 75% potassium nitrate, 15% charcoal and 10% sulphur. The explosive reaction is 10KNO3 + 3S + 8C (3K 2SO4) + 2K 2CO3 + 6CO2 ↑ + 5N2↑ If excess of carbon and sulphur take part in slower processes, it leads to evolution of more gases. It is an excellent and a cheap explosive for blasting down of coal, as its low velocity gives it a slow heating action that does not shatter the coal unduly. Hence these are known as time in delay-fuses; used for blasting in shells, igniters and primer assemblies for propellants, practice bombs and saluting charges. Nitrocellulose It is prepared by treating cellulose with nitric and sulphuric acids. Formed nitrocellulose is dissolved in a mixture of ether and alcohol and the solvent is evaporated, leaving a jelly-like solid behind. It is study by adding stabiliser like diphenylamine and pressed into cylindrical rods. It is called smokeless powder as it produces smokeless gases such as CO2, CO, N2 and water vapour.

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7.38 Engineering Chemistry High Explosives Explosives which have high energy content than the primary explosives are called high explosives. These are quite stable and are insensitive to fire, mechanical shocks, etc. Hence, to start a rapid chemical reaction, some amount of primary explosives are placed with high explosives. These are broadly divided into four groups: Single Compounds Explosives These contain only one chemical compound. Some of examples are as follows: (i) Ammonium nitrate: It is very stable, nontoxic and cheap and has low brisance value. It is about half as powerful as TNT and is mostly employed in making binary explosives. It is dangerous to store near any inflammable material. (ii) TNT (Trinitrotoluene): It is high explosive. It is most widely used in shell-firing and under-water explosions and is well-suited for loading in containers, due to low melting point. It is a safe explosive in manufacture, transportation, storage, non-hydroscopic and violent disruptive explosive, and does not react with metals. Hence, it is used for military purposes. (iii) PETN (Pentaerythritol tetra nitrate): It is an extremely powerful, sensitive and standard military explosive. (iv) RDX or cyclonite: It is a powerful, sensitive, less toxic, high explosive. RDX came into prominence in military as well as an industrial explosive. Binary Explosives Binary explosives are a mixture of TNT and other explosives, and these are more convenient and superior than single explosives. Due to their low melting point, TNT is one of the ingredients in all binary explosives. For example, (i) TNT + Ammonium nitrate is amatol. (ii) TNT + PETN is pentolite. (iii) TNT + Tetryl is tetrytol. (iv) TNT + RDX + Al powder is tropex. (v) TNT + Al flakes is titronal. Plastic Explosive These are a combination different explosives which are in plastic state. They can be hand moulded and made into various shapes without any serious risk and are mainly used for industrial applications and military uses. With high explosive simple combination of plastic explosive can prepare. Due to their engineering applications, they are available as flexible sheets. Dynamites These are explosives containing nitroglycerin as the main ingredient, by pressure or shock detonates. 4C3 H(NO)3 →10H 2 O + 6N ↑ + 12CO ↑ + O ↑ 1 Volume up to 10,000 volumes at temperature explosion

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The explosion is so sudden that it would shatter the breech of the rifle, before the bullet had time to move. It also pulverises the rock, instead of breaking it into fragments of immovable size. It is dangerous in handling and transporting and is usually mixed with an insert absorbent like wood pulp, starch, meals, etc. Some examples are as follows: (i) Straight dynamites: They contain 15–60% nitroglycerine in wood meal with sodium nitrate. The important uses are blasting of hard rocks, coal and other minerals. (ii) Blasting gelatin-dynamites: It is the combination of nitroglycerene (91.5%) partly gelatinised by nitro cotton (8%) or colladion cotton and CaCO3 (0.5%). These are very powerful, water-proof and stick well in holes into which they have been loaded. They can be used under wet conditions, where high loading density is desired. The important uses are submarine blasting, tunnel driving, deep-well shooting and at places where maximum shattering effects are desired. (iii) Gun cotton (or cellulose nitrate): Cotton is steeped for half an hour in a cooled mixture of concentrated nitric acid and concentrated sulphuric acid. C6H7O2(OH)3 + 3HNO3(C6H7O2(NO3)3 + 3H2O (iv) Cordite: It is a form of smokeless powder made by dissolving guncotton (65 parts), nitroglycerine (30 parts) and petroleum jelly or vaseline (five parts) in acetone. The resulting paste is rolled and cut into pieces of different dimensions, according to the rate of explosion desired. When acetone evaporates, the horny cordite remains. The vaseline acts as a stabiliser and a cooling agent on the powder as it has a tendency to lower the temperature of explosion. The guncotton slows down the explosive reaction of nitroglycerine and makes cordite an excellent propellant for large calibre naval guns. (v) Gelignite: It consists of 65% blasting gelatine and 35% of absorbing powder. It is a powerful explosive, which can be used under water.

7.6.3 Precautions during Storage of Explosives The following precautions should be taken while storing explosives: (i) Explosives should be stored separately, that is, without mixing with others. (ii) Detonators and explosives should be stored separately. (iii) Flame lanterns should not be used. In case of power failure, only torches should be used for lighting purposes. (iv) All electric fittings and wirings should be properly insulated and frequently checked. (v) Only authorised persons should be allowed to the explosive store by wearing magazine shoes. (vi) One should take care of jerks or drops while handling explosives. (vii) Smoking/fire should be strictly prohibited around the explosive store. (viii) The magazine (explosive store) should not be constructed within 500m from any working kiln or furnace. (ix) The boundary of the explosive store should be protected by high barbed wire fencing and proper caution boards should be put up. (x) The magazine should be provided with heightening conductors as a safeguard. (xi) During thunder, the magazine should not be opened and no one should be allowed inside.

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7.40 Engineering Chemistry

7.6.4 Blasting Fuses A fuse is “a thin water-proof canvas length of tube containing gunpowder (or TNT) arranged to burn at a given speed for setting off charges of explosives. Fuses are of two types as follows. Safety Fuse It is employed in initiating caps, where electrical firing is not used. A safety fuse consists of a small diameter core of black powder, enclosed in a covering of wrapper of water proofed fabrics. It is made to have an approximate burning speed of 30–40 seconds per foot (or about 1 cm s-1). When a fuse is used to fire a shot in blasting, sufficient length is used so that ample time is allowed for the shot firer to reach a point of safety. Detonating Fuse It has a velocity of over 6,000 m/s and consists of a charge of high velocity explosive, such as TNT, contained in a small diameter bent tube. The line of fuse (called Cordean) is in contact with the charge throughout its length and this causes practically instantaneous detonation of the whole charge, regardless of its velocity. Such fuses are used principally for exploding charges of explosive in deep holes.

7.6.5 Important Explosives and their Preparation (i) Lead AzidePb(N3)2 is prepared by reacting sodamide (formed by reacting ammonia with sodium) with nitrous oxide (N2O), followed by treating the sodium azide (NaN3) so-formed with lead acetate. Thus; 6NH3 + 2Na (2NaNH2 + 3H2) NaNH2 + N2O → (NaN3 + H2O) 2NaN3 + Pb(OAc) → Pb(N3)2 + 2AcONa Sodium azide Lead acetate Lead azide Sodium acetate (ii) Mercury Fulminate Hg(ONC)2 is prepared by dissolving mercury in excess of HNO3 and then pouring the resulting solution of ethyl alcohol, when the solution starts boiling and mercury fulminate is precipitated. Decomposition reaction: Hg(ONC)2 Hg(g) + N2 + CO + 117 k Cal (iii) Diazodinitrophenol (DDNP) is prepared by diazotising picranic acid. OH O2N

OH NH2

NO2

Diazotization

O N2Cl

O2N

−HCl

NO2

O2N

H N

NO2 4,6- dinitro-1,2 -diaxophenol (DDNP)

(iv) Trinitrotoluene (TNT) is prepared by the nitration of toluene (C6H5CH3) using a nitrating mixture of concentrated HNO3 and concentrated H2SO4 in 1:1 ratio in a tank reactor, in which contents are continuously stirred.

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CH3

CH3 (conc)

+ 3HNO3 (conc)

NO2

O2N

H2SO4

Toluene

NO2 TNT

The liquid product (TNT) so formed is taken out, washed with ammoniacal solution of Na2SO3 and then with cold water, when TNT crystallises out. Crystals of TNT are filtered and purified by melting. The melt is dried (by passing warm air) and poured in containers. Decomposition reaction: TNT (3.5CO + 3.5C + 2.5H2O + 1.5N2 + 190 K cal (v) Nitroglycerine (NG) or glycerol trinitrate (GTN) is prepared by adding glycerol to a cooled mixture of concentrated H2SO4 (60%) and concentrated HNO3 (40%) at 10°C with constant stirring. CH2OH CH2OH + 3HNO3 (conc)

CH2ONO2

conc H2SO4

CH2OH

CHONO2 CH2ONO2

After nitration, the mixture is run into a tank, when the nitroglycerine rises to the top, while excess acids form the lower layer. The nitroglycerine layer is separated, washed first with water and then with dilute sodium carbonate solution (2%) to remove traces of acids completely. It is then converted into different desired types of dynamites by absorbing in specific inert materials. Decomposition reaction: 4C3H5(NO3)2(12CO2 + 6N2 + 10H2O + O2 (vi) Pentaertythrital Tetranitrate (PETN) is prepared by the cannizaro reaction between formaldehyde and acetaldehyde in the molar ratio 4:1. CH2ONO2

CH2OH CH3CHO + 4HCHO

Ca(OH)2

HOH2C

C CH2OH

CH2OH

3HNO3 (conc) H2SO4 (conc)

O2NOH2C

CH2ONO2

CH2ONO2 PETN

Decomposition reaction: C(CH2ONO2)4(3CO2 + 2CO + 4H2O + 2N2 + 180 K Cal (vii) RDX or Cyclonite (Cyclomethylene trinitro amine) is prepared by treating hexamethylene tetra amine with nitric acid. (CH 2 )6 N 4 + 4HNO3 → (CH 2 NNO2 )3 + NH 4 NO3 + 3HCHO Hexamethylene tetra amine Formaldehyde

7.6.6 Rocket Propellants A rocket propellant is a high oxygen-containing fuel plus oxidant whose combustion takes place in a definite and controlled manner with the evolution of a huge volume of gas. A propellant reacts quickly, producing a very large volume of hot gases (usually at a temperature of 3,000 °C and a pressure of 300 kg/cm2), which exits through a small opening (called “jet” or nozzle) at supersonic velocity. This act of pushing the gas backwards produces an equal and opposite reaction—Newton’s third law of

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Kerosene (C12H26)

Liquid oxygen (O2)

motion, which moves the rocket forward. With the increase of exhaust velocity, the rocket performance increases. Rockets are used for pyrotechnic effect signalling, carrying a life, hurling explosives at an enemy, putting space capsule into orbit, etc. (Figure 7.20).

C12H26 + 18.5 O2

Injection Chamber

12CO2 + 13H2O

Combustion Chamber

Figure 7.20 Use of bipropellant in a rocket. Combustion reaction between kerosene and liquid oxygen produces a huge volume of gaseous CO2 and H2O, the thrust of which forces the rocket upwards with a high speed

7.6.7 Characteristics of a Good Propellant The characteristics of a good propellant are as follows: (i) It should have high specific impulse (specific impulse is the thrust delivered divided by the rate of propellant (fuel plus oxidant) burnt. (ii) It should produce low molecular weight products (like H2, CO, CO2 and N2) during combustion. (iii) It should burn at a slow and steady rate. (iv) It should possess low ignition delay (ignition delay is the time taken by propellant to catch fire in the presence of an oxidising agent. It is expressed in milliseconds) (v) It should possess high density. (vi) It should be stable over wider ranges of temperature. (vii) It should be safe to handle and store under ordinary condition, that is, it should not detonate under shock, heat or impact. (viii) It should be non-corrosive and non-hydroscopic. (ix) It should be readily ignitable at a predictable burning rate. (x) It should leave no solid residue after ignition. (xi) It should not produce toxic products. (xii) It should produce high temperature on combustion. Note: Specific impulse is the thrust in kg per second on the fuel burnt. The value of thrust (or propulsive force) due to momentum of the exiting gases and is given by the following formula: F=

m v + ( Pe − Pa ) g

where F = thrust (kg/m/kg) m = mass flow (kg) g = acceleration due to gravity (9.81 m 2/s)

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v = exhaust velocity (m/s) Pe = exit pressure (kg/m 2) Pa = ambient gas pressure (kg/m 2) and Ae = nozzle exit area (m 2)

7.6.8 Classifications of Propellants The chemical propellants are classified into solid and liquid categories. Solid Propellants These may be “homogeneous” or “composite”. When a solid propellant or a mixture of propellants are thoroughly mixed in a colloidal state, it is called homogenous solid propellant. When a single propellant is employed, it is called a single-base propellant. For example, nitro cellulose, also known as guncotton or smokeless powder. On the other hand, a solid propellant which contains two materials is called a double-base propellant. For example, ballistite, containing nitro cellulose and nitroglycerine mixture, is a powerful double-base solid propellant. Cordite, composed of 65% nitrocellulose, 30% of nitro cellulose and 40–45% nitroglycerine. Diethyl phthalate, up to 5%, is also added and this acts as a solvent cum plasticiser, thereby giving a homogenous plastic mass, which can be worked smoothly. In addition, up to 1% of diphenylamine is added, which acts as a stabiliser. The propellant gives a flame temperature of about 2,700°C and the volume of gases is about 1,500 times the original volume. When an oxidising agent is dispersed in a fuel mass, the solid propellant is heterogeneous or composite. Gunpowder is the oldest composite propellant. It gives a flame temperature of 800–1,500°C and volume of gases is about 400 times the volume of the charge. Other composite solid propellants are 75% potassium per chlorate plus 25% asphalt oil, 80% ammonium perchlorate plus 20% resin (rubber) binder, 46% ammonium picrate, 48% sodium nitrate and 8% plastic resin binder. In selecting the oxidiser, it should be seen that it is non-hygroscopic, stable in contact with fuel and does not form any corrosive products. The oxidant potassium perchlorate leaves behind a white residue of potassium chloride particles; while ammonium perchlorate leaves no solid residue, but its combustion products contain hydrogen chloride and water, which form a toxic and corrosive fog. Liquid Propellants They possess many advantages over solid propellants. Thus, liquid propellants are more versatile and the engine using them can be checked and more easily calibrated. However, unlike solid propellants, the engine using liquid propellant is quite delicate and cannot withstand any rough handling. Liquid propellants may be monopropellants or bipropellants. Monopropellants It has fuel as well as oxidiser in the same molecule or in a solution containing both these. For example, hydrogen peroxide, nitromethane, ethylene oxide hydrazine propyl nitrate and a mixture of 21.4% methanol and 78.6% hydrogen peroxide are some of the common propellants (mono). A monopropellant must be safe to state and at the same time, it should burn smoothly. Due to its reactivity, hydrogen peroxide is not easy to store and handle. Moreover, metal oxides catalyse decomposition. Therefore, storage tanks are made of special materials for the same.

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7.44 Engineering Chemistry Bipropellants These are more widely used. Here, liquid fuel and oxidiser, kept separately, are separately injected in the combustion chamber. The commonly used fuels are liquid hydrogen, hydrazine, ethyl alcohol, aniline and kerosene oil. Ethyl alcohol admixed with 25% water is a good fuel. Although the addition of water reduces flame temperature, it reduces the molecular mass of combustion gases, which compensates for reduction in performance. The common oxidisers employed are liquid oxygen, ozone, hydrogen peroxide, fuming nitric acid and liquid flourine. Liquid oxygen is a non-toxic, safe and good oxidising agent, but it has to be stored under pressure in insulated containers. Ozone is a very powerful oxidising agent, but it is quite toxic and can explode at high concentration. Liquid flourine is volatile, toxic, corrosive, very reactive, but is a very good oxidising agent. It is also difficult to store and handle it.

7.7 NANOMATERIALS Nanoscience and technology are considered one of the most promising fields having huge potential to bring countless opportunities in many areas of research and development. It is the study of tiny structures at nanometer scale, which forms a basis for several core technologies. Definition “Nanotechnology plays a key role in many areas”. “One nanometer is one billionth of a meter”. 1 nm = 1/1,00,00,00,000 of a metre, which is close to 1/1,00,00,00,000 of a yard. For getting easy sense of the nanoscale, it is suffice to know that a human hair has around 50,000 nm and a commonly used microchip has around 150 nm. The normal human eye can see the things which have the size above 10,000 nm only. Note on Measures Almost all nanosciences are discussed by using SI measurements; SI Units and their description shown in Table 7.3. Table 7.3 Measuring units and their description SI unit

Description

Meter (m) Centimeter (cm) Millimeter (mm) Micrometer (μm) Nanometer (nm)

Approximately three feet or one yard. 1/100 of a meter, around half an inch (10-2 m) 1/1,000 of a meter (10-3 m). 1/100,000 of a meter (10-5 m), also called micron; this is the scale of the most integrated circuits. 1/1,00,00,00,000 of a meter (10-9 m). The size scale of single, small molecules and nanotechnology.

7.7.1 Synthesis of Nanomaterials Nanotechnology has sprung into prominence due to the recent development of various synthesis techniques/methodologies and the discovery of modern tools for the characterisation and manipulation of nanomaterials. Synthesis techniques are broadly categorised as top-up approach and bottom-up approach. Some of the methods for synthesis of nanomaterials are as follows: (i) Vapour–liquid–solid growth

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(ii) (iii) (iv) (v)

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Solution–liquid–solid growth Template–mediated growth Electron beam lithography Reverse micellar route, etc.

Preparation of Copper and Nickel Nanooxalates by Reverse Micelle Route The reverse micellar route is the best method for synthesis of variety of nanoparticles due to the ability to mix the reactants efficiently and control the size of nanoparticles effectively. The synthesis of these nanorods have been achieved using two micro-emulsions as described here: (i) Microemulsion I is the mixture of cetyl trimethyl ammonium bromide (CTAB) as a surfactant and n-butanol as the co-surfactant. Here, isooctane or n-octane is used as a hydrocarbon phase and 0.1 M copper nitrate/nickel nitrate solution is used as the aqueous phase. (ii) Microemulsion II comprises the same constituents as microemulsion I, except for having 0.1 m ammonium oxalate instead of copper nitrate or nickel nitrate as the aqueous phase. The weight fractions of various constituents in these microemulsions are 16.76% of CTAB, 13.90% of n-butanol, 59.29% of isooctane and 10.05% of the aqueous phase. These two microemulsions were mixed slowly and stirred overnight on a magnetic stirrer, and the resulting precipitate was separated from the apolar solvents and surfactants by centrifuging and washing it with 1:1 mixture of methanol and chloroform. The precipitate is then dried in air.

7.7.2 Characterisation The nano materials are characterised by X-ray diffraction, transmission electron microscopy, scanning electron microscopy, (TEM, SEM and AFM), dynamic light scattering studies (DLS), thermal analysis (TGA/DTA), Fourier transform infrared spectroscopy (FTIR), X-ray photoelectron spectroscopy (XPS), field dependent magnetisation studies, etc.

7.7.3 Importance (i) The synthesis of nanowires and nano rods has generated a lot of interest in the recent years due to their importance as objects for understanding microscopic systems. (ii) Nano-structured wires and rods are expected to have interesting optical, electrical, magnetic and mechanical properties as compared to micron-sized whiskers and fibres. (iii) There also find potential use as nanowires in the area of biosensors, where a specific molecule is attached to the tip of the nanowire, which then identifies a particular molecule in the living system.

7.7.4 Broad Classification of Nanomaterials According to the arrangements of atoms or molecules, nanomaterials are broadly classified into three types as follows: (i) Materials which have one dimension in the nanoscale. For example, surface coatings and thin films.

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7.46 Engineering Chemistry (ii) Materials which have two dimensions in nanoscale. For example, nanowires and nanotubes. (iii) Materials which have three dimensions in nanoscale or quantum dots. For example, fullerenes, cadmium-selenium quantum dots, gold quantum dots, ZnO quantum dots, etc. Some Other Nanomaterials Super-paramagnetic iron oxide nanoparticle, ZnO nanoparticle, titanium oxide nanoparticle, carboncoated silver nanoparticles, gold nanoparticle, NaYF4 nanophosphors, CdS nanowires, ZnS nanorods, Fe3O4 nanoparticles, etc., are some other nanomaterials.

7.7.5 Fullerenes A fullerene is a molecule of carbon in the form a hallow sphere, ellipsoid, tube and many other shapes. A spherical fullerene are also called Buckminsterfullerene (bucky balls); they resemble the balls used in football. The cylindrical one is called carbon nanotubes or bucky tubes. The fullerene is similar to structure graphite, which is composed of stacked graph sheets of linked hexagonal or pentagonal rings. Buckminsterfullerene (C60) was prepared in 1985 by Richard Smalley, Robert curl, James heath, Sean O’Brien and Harold Kroto. Kroto, Curl and Smalley were awarded the 1996 Nobel Prize in Chemistryfor their roles in the discovery of this class of molecules.

7.7.6. Types of Fullerenes The types of fullerenes are discussed in this section. Bucky Ball Cluster Since the discovery of fullerenes in 1985, many types of fullerenes like Bucky ball clusters, nanotubes, mega tubes, nano “onions”, linked “ball and chain” dimmers and fullerene rings are prepared. The most common bucky ball is C60 as shown in Figure 7.21.

Figure 7.21 Common buck ball Buckminsterfullerene was named after Richard Buckminster Fuller, a noted architectural modeller who popularised the geodesic dome. As buckminsterfullerenes have a shape similar to that sort of a dome, the name seemed to be appropriate.As the discovery of the fullerene family cameafterbuckminsterfullerene, the shortened name “fullerene” is used to refer to the family of fullerenes. The suffix “-ene” indicates that each C atom is covalently bonded to three others (instead of the maximum of four),

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a situation that classically would correspond to the existence of bonds involving two pairs of electrons. Buckminsterfullerene is the smallest fullerene molecule containing hexagonal and pentagonal rings, in which two pentagons share on edge. It is also the most common is terms of natural occurrence, as it can often be found in soot. The structure of C60 is a tralancated icosahedron, which resembles an association football of the type made of 20 hexagons and 12 pentagons with carbon atom at the verities of each polygon and a bond along each polygon edge. The Van der Waals diameter of a C60 molecule is about 1.1 nm. The nucleolus to nucleolus diameter of α C60 molecule is about 0.71 nm. The C60 molecule has two bond lengths. Applications Fullerenes have been extensively used for several bio-medical applications including the design of high performance MRI contract agents, X-ray imaging contract agents, photo dynamic therapy and drug and gene delivery and have been summarised in several comprehensive reviews. Carbon Nanotubes Nano tubes are cylindrical fullerenes. These tubes of carbons are usually only a few nanometers wide, but they can range from less than micrometer to several millimeters in length. They often have closed ends but can be open-ended as well. Their unique molecular structure results in extraordinary microscopic properties including high tensile strength, high electric conductivity, high ductile, high heat conductivity and relative chemical in actively. Hallow tubes of very small dimension with single or multiple walls having potential application in electronics and industry. Carbon nanotubes have been identified into three structures namely, armchair, zigzag and chiral structure (Figure 7.22 (a), (b) and (c)).

(a)

(b)

(c)

Figure 7.22 (a) Armchair (n, n), (b) Zigzag (n, 0), (c) Chiral (n, m) Due to well-defined geometry, exceptional mechanical properties and extraordinary electrical characteristics of carbon nanotubes (CNTs) are used as nanoelectric circuits, nanoelectrochemical systems, nonorobots, etc.

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7.48 Engineering Chemistry Synthesis of Carbon Nanotubes Carbon nanotubes, including powder and array types, are generally produced by three main techniques. (i) Electric arc discharge method (ii) Laser ablation method (iii) Thermal or plasma-enhanced chemical vapour deposition method Laser ablation and arc discharge methods are modified physical vapour deposition techniques and involve the condensation of hot, gaseous carbon atoms generated from the evaporation of solid carbon. In arc discharge, vapour is created by an arc discharge between two carbon electrodes with or without a catalyst. The nanotube self-assemble from the resulting carbon vapour laser ablation produces a small amount of clean nanotubes, whereas arc discharge methods generally produce large quantities of impure material. Laser ablation and arc discharge techniques produce powder-type nanotubes with impurities in the form of amorphous carbon and catalyst particles because of the high temperature of the heat source. The chemical vapour deposition method is a better technique for high yield and low impurity production of carbon nanotube arrays at moderate temperatures. Plasma-enhanced chemical vapour deposition can grow individual, free-standing nanofibers with special controlling ability. On the other hand, high-density aligned nanotubes can be mass-produced using thermal chemical vapour deposition. The method also provides good control over the size, shape and alignment of the nanotubes. Recently, Kang et al. have synthesised dense, perfectly aligned arrays of single-walled nanotubes, and Vijayaraghavan et al. have been able to horizontally align individual nanotubes on a large scale using dielectrophoretic force. Nanotube materials for analytical applications should be clean with low impurities from metal catalysts and amorphous carbon. The synthesis of nanotubes with a uniform size and special density is critical before post-processing and functionalisation. Applications of Carbon Nanotubes (i) Carbon nanotubes are used in the preparation of nanoelectric circuits, nanoelectromechanical systems and nanorobotic systems. (ii) Multi-walled carbon nanotubes can serve as bearings, switches, gigahertz oscillators, memories, shuttles, syringes, etc. (iii) The hollow carbon nanotubes can be used as containers, conduits, pipettes, coaxial cables, etc. Carbon Nanotubes as Nano Biosensors Carbon nanotubes have become the forces of intensive research by analytical chemists, as electrodes to transmit electrical signals or as sensors to detect concentrations of chemicals, biological materials. Fascinating physical and chemical properties such as electrical conductance, high mechanical stiffness, light weight, electron-spin resonance, field emission, electrochemical actuation, transistor behaviour, piezoresistance, contact resistance, coulomb drag power generation, thermal conductivity, luminescence, electrochemical bond expansion, opto-mechanical actuation and the possibilities of functionalising carbon nanotubes to change their intrinsic properties are the reasons for their use as novel biosensors. The structure of nanotubes can be described as a rolled-up tabular shell of graphite sheet with the carbon atoms covalently bound to their neighbours. The helicity of the shell categorises nanotubes

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into metallic or semiconducting types. Based on the capture and promotion of electron transfer reaction from analytics, ballistic conductivity of metallic nanotubes is extremely attractive. Semiconducting nanotubes can be used as biosensors directly because these are optimised by changing the gate voltage. Hence, CNT electrodes have high sensitivity with low detection limit. CNT can be described as single wall nanotubes (SWNT), double wall nanotubes (DWNT) or multi-wall nanotubes (MWNT). CNT diameters range from about 1.4 nm to 60 nm and their length varies from microns to above one centimeter. To use nanotube electrodes for electro analytic purposes, proper conjugation strategies between biological molecules such as enzymes, single stand DNA/RNA/PNA, antibodies, receptors and aptamers need to be developed. Appropriate functionalisation methods and immobilisation of biomaterials on nanotubes are critical since functional groups create defects in the nanotubes that will eventually alter or degrade the intrinsic electrical properties of the nanotube. Nanotubes also have contact resistance that should be minimised to provide the highest sensitivity when used as a sensor. Carbon Nanotubes as Nanorobotics Robots can do all kinds of physical and mechanical work more in various situations instead of manpower. Nowadays, large-sized robots are used, and shrinking the robot size to nanoscale with carbon nanotube is a fascinating advantage. Nanorobots can measure mass in femtogram ranges, sensing forces at piconewton scales, etc., and with a provision for advanced features nanorobots can introduced in army, intelligence, security, mechanochemical synthesis, etc.

7.7.7 Properties of Nanomaterials The advent of nanotechnology has resulted in the increased use of nanomaterial-based products in daily life. A significant increase in the surface-area-to-volume ratio at the nanoscale, giving rise to novel and enhanced magnetic, mechanical, electronic, catalytic, conducting and optical properties of nanomaterials, has made nanotechnology the most promising tool of this century. Size Nanostructures are the smallest of human-made things but the largest molecules of natural things and nanometer is the magical point on the scale. Nanoscience and technology provide advanced materials and systems which are intermediate between isolated atoms and bulk materials, with controlling of transitional properties. Due to drastically increased surface-to-volume ratio, physical and chemical properties are quite different from the bulk but tend to dominate at nanoscale. Optical Properties Due to large surface area, nanomaterials exhibit better optical properties. Luminescent nanocrystals, termed as nanophosphors, with unique optical properties make them ideal for a wide spectrum of applications ranging from flexible displays, lasers to biological imaging and therapeutic agents. For example, NaYF4 is a highly multifunctional material with promising potential application in IR to visible up conversion process. Silicon nanowires show strong photoluminescence.

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7.50 Engineering Chemistry Mechanical Properties The mechanical behaviour of nanomaterial are found to be high strength, good ductility, superior superplasticity, a low-friction coefficient, good thermal stability, high wear resistance, enhanced highcycle fatigue life and good corrosion resistance. For example, tensile strength of carbon nanotube is approximately 20 times greater than the steel. Magnetic Property The magnetic property of a nanomaterial is very useful in biological systems. Due to unique properties and biocompatibility of nanoparticles, they easily combine with biological materials like proteins, antibodies, enzymes, nucleus acids, etc. Electrical Properties Nanomaterials show electrical conductivity from semiconductors to superconductors, depending upon the diameter and chirality of the molecules. Nano-structured conducting polymers have generated much interest for their potential use in nanoelectronics. For example, fullerenes are a class of allotropes of carbon, which are basically graphene sheets rolled into tubes or spheres. They include the carbon nanotubes because of their mechanical strength and electrical properties. Carbon nanotubes have higher electrical conductivity than copper wires. The high electrical conductivity of nanotubes is mainly negligible amount of defects, so they possess low resistance. Semiconductors Most nanomaterials show semi-conductivity. For example, carbon nanotubes, nanowires, MoS2, etc. They are used mainly as field-effect transistors, p-n diodes, etc. Superconductors Some nanomaterials at low temperature show superconductivity. For example, at normal temperature, NbS2 is metallic in nature, but at low temperature, it becomes a superconductor. Catalytic Activity Due to the increasing surface area, nanomaterials act as a good catalyst in different homogeneous and heterogeneous phases. (i) For example,In 2O3-Ga2O3/Al2O3 nanocomposite used in selective catalytic reduction of nitric oxide. (ii) Platinum, a precious transition metal, which has outstanding catalytic and electrical properties and superior resistant characteristics to corrosion, has been widely applied in chemical, petrochemical, pharmaceutical, electronic and automotive industries. Both platinum metal and

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its alloys possess distinctive ability in catalysing partial oxidation, hydrogenation and dehydrogenation of a variety of important molecules that are essential in many industrial processes.

7.7.8 Applications of Nanomaterials Due to the advent of nanotechnology, applications of nanomaterials are quickly increasing by the day. (i) Conducting polymers are widely employed as coatings conferring electrode systems, antifouling properties and possibly activating electrocatalytic redox processes, conducting polymers for commercial applications due to their high conductivity, easy preparation and environmental stability have generated much interest for their potential use in nanoelectronic and organic conductors. (ii) Aurum nanoparticles have been used for the enhancement of radiation effects on bovine aortic endothelial cells of superficial X-ray radiation therapy and megavoltage electron radiation therapy beans. (iii) Nanoparticles are used as important catalysts in different chemical reactions. (iv) Quantum dots and quantum wires have been used in the design of new super computers. (v) Magnetic nanoparticles have wide application in medicine as drug transport and biosensor. (vi) Carbon nanotubes having fascinating physical and chemical properties such as electrical conductance, high mechanical stiffness, light weight, electron spin resonance field emission, electrochemical actuation, transistor behaviour, piezoresistance, contact resistance, coulomb drag power generation, thermal conductivity, luminescence, electrochemical bond expression, optomechanical actuation,and the possibilities of function-aligned CNTs to change their intrinsic properties are the reasons for their use in different ways as follows: (a) They are used as a novel biosensor as enzyme electrodes, immunosensors, DNA, RNA, electrodes, etc. (b) Due to well-defined geometry, exceptional mechanical properties and extraordinary electric characteristics, among other outstanding physical properties of CNTs, they are qualified for potential applications in the preparation of nanoelectronic circuits, nanoelectromechanics systems, nanorobotics systems, structural elements, probes, grippers, tweezers, scissors, actuators, bearings, syringes, switches, memories, diodes, transistors, logic gates, wires, cables, position sensors, displacement sensors, circuits, thermal actuators, etc. (c) They are used in field emission light devices for fluorescent displays. (d) They act as a storage device in lithium batteries. (e) From household uses to industries, space, army, medicine, agriculture and other fields, nanotechnology applications are employed.

7.8 LIQUID CRYSTALS The study of liquid crystals began in 1888 with the observation of two distinct melting points to cholesteryl benzoate by Friedrich Reinitzer. Liquid crystals constitute a fascinating state of aggregation that is intermediate between the crystalline solid and the amorphous liquid. They exhibit rheological behaviour similar to those of liquids and anisotropic physical properties similar to crystalline solids. Their dual nature and easy response to electric, magnetic and surface forces have generated innumerable applications, which are extended

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7.52 Engineering Chemistry over diverse fields. The explosive growth of application of liquid crystals has generated immense interest amongst the scientists and technologists to synthesise, characterise and understand the substances. “Liquid crystal is formed in between the temperature of transition point and melting point, the cloudy liquid shows double refraction. Hence the cloudy liquid is known as liquid crystal or mesomorphic state”. “The liquid crystals are highly anisotropic fluids that exist between the boundaries of the solid and conventional, isotropic liquid phase”. Liquid crystals are made up of liquid and crystals. The term “liquid” is used because the tendency to take the shape of the container and the word crystals is used because they still contain one or two dimensional arrays.

7.8.1 Characteristics of Liquid Crystal Phase The liquid crystalline state is the tendency of the mesogens (molecules) to point along the director (common axis), which have no intrinsic order. The average alignment of the molecules in solid, liquid crystal and liquid are shown in Figure 7.23. Crystalline material possesses long range periodic order in three dimensions, but liquid crystals do not possess order as a solid, since they have only some degree of alignment order. They show positional, orientational, bond orientational order, etc. These are characterised by partial ordering, that is, one or more degrees of freedom, but not all will have the long-range order.

Solid

Liquid crystal

Liquid

Figure 7.23 Average alignment of the molecules in solid, liquid crystal and liquid

7.8.2 Classification of Liquid Crystals Liquid crystals may be divided into two broad categories, according to the principal means of breaking down the complete order of the solid state. (i) Thermotropic liquid crystals (ii) Lyotropic liquid crystals

7.8.3 Thermotropic Liquid Crystals These crystals are established solely by the adjustment of temperature. There are divided into three types, namely nematic, cholesteric and smectic. These three classes are distinguished by the different kinds of molecular order they exhibit.

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Nematic Structure The molecules in the nematic structure maintain a parallel or nearly parallel arrangement to each other along the long molecular axes. They are mobile in three directions and can rotate about one axis. The schematic representation of nematic liquid crystals are shown in Figure 7.24. Properties The nematic structure is the highest temperature mesophase in thermotropic liquid crystals. In this structure, the molecules have no particular positional order, but tend to point vertically in same direction as shown in Figure 7.24. Like liquids, molecules are free to flow, their centre of mass positions are randomly distributed, but they maintain long range directional or orientational order. The molecular order of nematics is shown schematically in Figure 7.24. As in any liquid, the molecules possess no translational order. There exists, however, a significant degree of long-range orientational order.

^ n

Figure 7.24 Schematic representation of molecular order in nematic liquid crystals Cholesteric Structure The cholesteric-mesophase is a nematic type of liquid crystal except that it is made up of optically active molecules. The cholesteric liquid crystal phase is typically composed of nematic-mesogenic molecules containing a chiral centre, which produces intermolecular forces that favour alignment between molecules at a slight angle to one another. Helical structure and pitch of helix of the cholesteric liquid crystal are shown in Figs. 7.25 (a) and (b) respectively.

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7.54 Engineering Chemistry Molecules Sheets

Figure 7.25 (a) Helical structure of the cholesteric liquid crystal (b) Pitch of a helix of cholesteric liquid crystal Properties The cholesteric structure acquires a spontaneous twist about an axis normal to the preferred molecular directions. The twist may be right-handed or left-handed depending on the molecular conformation. In the structure of cholesteric phase, the local molecular ordering is identical to that of the nematic phase. The chiral nematic phase exhibits chirality, hence it is called cholesteric phase. Chiral molecules those have no internal planes of symmetry can give cholesteric phase. This phase shows a spontaneous twisting of the molecular axis parallel to the director. This twist may be right- or left-handed and depends on the molecular conformation. In the structure of cholesteric phase, the local molecular ordering is identical to that of the nematic phase. Due to asymmetric packing, the finite twist angle between adjacent molecules is a long range chiral order. A cholesteric liquid crystal rotates the direction of linearly polarised light, and it is roughly 1,000 times stronger than the activity of an ordinary optically active substance such as quartz, hence it is optically active. The chiral pitch refers to the distance over which the liquid crystal molecule undergoes a full 360-degree twist. Smectic Structure The word “smectic” is derived from the Greek word which means soap. The origin is explained by the fact that the thick, slippery substance often found at the bottom of a soap dish is actually a type of smectic liquid crystal. Smectic liquid crystal is shown in Figure 7.26.

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Properties Another distinct mesophase of liquid crystal substance is the smectic state. In this phase, the molecules show a degree of translational order which is not present in the nematic structure. In the smectic state, molecules maintain the general orientational order of nematics, but also tend to align themselves in layers or planes; the motion is restricted to within these planes, and separate planes are observed to flow part each other. The increased order means that the smectic state is more “solid-like” than the nematic state.

Figure 7.26 Smectic liquid crystal On the basis of appearance under a polarising microscope, the miscibility with known phases and X-ray scattering up to nine thermotropic smectic phases has been identified; these phases are labelled by the chronological order of their discovery and designated as smectic A, B … I. Some thermotropic liquid crystalline compounds are given in Table 7.4. Table 7.4 Thermotropic liquid crystalline compounds Name

Formula

Liquid crystalline range (°C )

1. Nematic Liquid Crystals (a) P-methoxybenzylideneP1N-butyl-aniline H3 C (b) P-n-Hexyl-P1-cyano Biphenyl

21−47

H13C6

C4H9

n 14−28

(Continued )

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7.56 Engineering Chemistry Table 7.4 (Continued) Name

Formula

Liquid crystalline range (°C )

2. Cholesteric Liquid Crystal: Cholesterylnonanoate

CH3 CH3

145−179

CH3

C (CH2)3 CH H

CH3

O H3C (CH2)7 C

3. Smectic Liquid Crystals: Smectic A: Ethyl P-(P1-Phenyl benzalamine) benzoate Smectic B: Ethyl P-ethoxy benzalP1-aminocinnamate

121−131

H O

COOC2H5 77−116

H H5C2O

Smectic C: P-n-octyloxy benzoic acid

H17C8

Smectic D: P1-n-Octadecyloxy-31-nitro diphenyl-P-Carbonylic acid

H37C18

COOC2H5

108−147

COOH

159−195

COOH

O2N Smectic E: Diethyl p-terphenyl-P, P11-Carbenylate

C2H5OOC

Smectic F: 2-(P-Pentyl Phenyl)5-(P-Pentyloxy Phenyl) Pyrimidine

Smectic G: 2-(P-Pentyl Phenyl)5-(Pentyloxy phenyl) Pyrimidine

Smectic H: 4-Butyloxybenzal-4ethyl Aniline

173−189

COOC2H5

103−114

N H11C5

H11C5

C5H11

79−103

N O

C5H11

40.5−51

H C4H9

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7.8.4 Lyotropic Liquid Crystals Mixtures of two or more components that change phase with changes of concentration are called lyotropic. These mesophases occur in concentrated solutions of rod-like molecules in an isotropic solvent. The stability of these mesophases is as readily influenced by concentration of solute as by temperature. Lyotropic mesophases are important in soaps, gels and colloids and are of great interest in biology. Lyotropic crystal that has one of the components is an amphiphile (containing polargroup attached to long hydrocarbon chains) and another is water. When a crystalline amphiphile is added to water, several mesophases can be observed, ranging from a true solution to the crystal state. For example, sodium stearate CH3(CH2)16COO –Na CH2

(CH2)16

CH3 CH3

O−

a-Lecithin-

CH2

(CH2)16

CH2

N+(CH3)3

7.8.5 Chemical Properties of Liquid Crystals According to chemical properties, liquid crystals can be broadly classified into two categories—thermotropic liquid crystals and lyotropic liquid crystals. These two are similar in many ways and are distinguished by the mechanisms that drive their self-organisation. Thermotropic liquid crystals are formed by increasing the temperature of a solid or by decreasing the temperature of a liquid. Usually, thermotropic liquid crystal occurs with anisotropic dispersion forces of the molecules and their packing interactions. Generally, thermotropic liquid crystals occur because of the presence of anisotropic dispersion forces between the molecules and packing interactions. According to their properties, thermotropic liquid crystals can be classified into two types. Enantiotropic Liquid Crystals Crystals that can be changed into the liquid crystal state from either lowering the temperature of a liquid or raising of the temperature of a solid are called enantiotropic crystals. Monotropic Liquid Crystals Crystals that can only be changed into the liquid crystal state from any one of these but not both by either an increase in the temperature of a solid or decrease in the temperature of a liquid are called monotropic liquid crystals. Lyotropic liquid crystals are formed with the influence of solvents, and not by changing of temperature. They occur with the results of solvent-induced aggregation of the constituent mesogens into micellar structures. Such crystals show amphiphilic properties, due to presence of both lyophilic and lyophobic ends. With this property, in the presence of a solvent form as micellar structures, because lyophobic ends will stay together and the lyophilic ends stay outward of the solution. By increasing the concentration of solution and cooling the size of micellar structures increase eventually and separate the newly formed liquid crystal from the solvent.

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7.8.6 Applications of Liquid Crystals Liquid crystal science had a major effect in different fields of science and technology. It is widely used in research, medicine, display, radiation-sensors, thermometers, non-destructive testers, etc. Displays These are more desirable for displays than other materials due to lower power consumption and the clarity of display in the presence of bright lights. The power requirements are low for digital display and other mechanism to runs the large watches. The two most widely used displays in liquid crystal displays are dynamic-scattering and field-effect crystals. Radiation and Sensors Cholesteric liquid crystals have been used in versatile and inexpensive radiation sensors, where an impinging invisible radiation is registered as a colour change by local heating and the change of structure. These devices can be used with UV, infrared, microwave, ultrasonic, ionizing radiation transducers, etc., where there is a practical absorber in contact with the liquid crystals. Cholesteric liquid-crystal films have been used as recording media in holograms. Thermometers The temperature-dependent variation in the colour of cholesteric liquid crystals has led to the use of these substances in the measurement of temperature and gradients of temperature. A cholesteric substance or a mixture of cholesteric substances always exhibits the same colour at the same temperature; the colour is very sensitive to change in the ambient temperature. Cholesteric liquid-crystal substances, when applied to the surface of the skin, have been used to locate veins, arteries, infections, tumours and the foetal placenta which are warmer than the surrounding tissue. Research Field Nematic liquid crystals are the most useful research tools in the application of magnetic resonance. Molecules that are dissolved in nematic liquid crystals due to anisotropic environment give a very highly reduced nuclear magnetic resonance (NMR) spectrum. The analysis of spectra of molecules in liquid crystal solvents yields information regarding the anisotropy of chemical shifts, chemical structure, bond angles, bond lengths, direct magnetic dipole-dipole interactions, indirect spin-spin couplings, molecular order and relaxation processes. Some liquid crystals have been used in chromatographic separations, as solvents to direct the course of chemical reactions and to study molecular rearrangements and kinetics and as anisotropic host fluids for visible, UV/IR spectroscopy of molecules. Non-destructive Testing Cholesteric-nematic liquid crystals are colour-sensitive with temperature. This property can be used for non-destructive testing. Biological Systems and Medicine Liquid crystals are used in medicine, as optical discs, full colour “electronic slides”, light modulators, etc. Most biological systems exhibit the properties of liquid crystals. Considerable concentrations of

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mesomorphic compounds have been found in many parts of the body, often as steroid or lipid derivatives. A liquid crystal phase has been implicated in degenerative diseases; for example, atherosclerosis, sickle-cell anaemia, etc. Living tissues such as muscle, tender, ovary, adrenal cortex, nerve etc., show the optical birefringence properties that are characteristic of liquid crystals.

7.9 ABRASIVES A material or a mineral that is used to shape, polish or finish a work surfaces through rubbing is an abrasive. These are substances characterized by their hardness, and they are used to wear down softer surfaces by cutting, grinding or polishing. Abrasives are widely and very extensively used in a wide variety of industrial, domestic and technological applications. This gives rise to a large variation in the physical, chemical composition, as well as the shape of the abrasive. Common uses for abrasives include grinding, polishing, cutting,drilling, sharpening, sanding,etc. Examples: (i) Housewives and farmers use abrasive stones to keep their kitchen knives and agricultural instruments sharp. (ii) Dentists use an abrasive powder when they clean teeth and to smooth down fillings. (iii) Abrasives play important roles in various industries such as in grinding of wood into paper pulp, cutting of stone into carved and surfaced structures, and sharpening of cutting-tools. Abrasives are broadly divided into natural and artificial abrasives. Some natural abrasives are calcite, emery, diamond, pumice, sand, corundum, garnet, etc. Artificial abrasives include ceramic, corundum, glass powder, silicon carbide, zirconia alumina, etc.

7.9.1 Hardness of Abrasive The most important quality of an abrasive is its hardness; it is measured roughly on Moh’s or Vicker’s scale as shown in Figure 7.27. Commonly used natural and artificial abrasives in order of increasing hardness are as follows: Talc < gypsum or NaCl < calcite < fluorite < apatite < feldspar < quartz < topaz or emery < corundum < diamond Generally, artificial abrasives are superior in uniformity than natural abrasives, and consequently, they are preferred in industrial grinding, etc. Moh’s scale of Hardness

Talc

Calcite

Gypsum

Apatite

Fluorite

7 Quartz

Feldspar

Corundum Topaz

Diamond

Figure 7.27 Moh’s scale of hardness

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7.9.2 Natural Abrasives Some of the natural abrasives are as follows: (i) Diamond: It is a crystallised carbon having highest hardness; hence, it can cut into the surface of any other substances, whereas only another diamond can scratch with a diamond. It is chemically highly inactive and not attacked by acids, alkalis or fused potassium chlorate. Diamonds that are off-colour or otherwise faulty are called borts. Black diamond from Brazil is called carbonado and has no value for jewellery. Borts and carbonado are used in drill points as saw-teeth for cutting rocks, stones, or grinding wheels, dressers, and in operations, where great hardness is required. (ii) Corundum: It is crystallised aluminium oxide (Al2O3) and comes next to diamond in hardness. Corundum is not of gem quality but finds use as abrasives for grinding glasses, gems, lenses, metals and metal-cutting. (iii) Emery: Emery is fine-grained, opaque, massive mineral and dark grey to black in colour. It is found mainly in Greece. It has an aggregate of 55%–75% crystalline alumina, 20%–40% magnetite (Fe3O4) and about 12% of other minerals of which the chief is tourmaline. Its hardness is about eight on Moh’s scale. The grinding power of emery almost depends on the proportion of alumina it contains and on the effects of the other ingredients in determining true hardness of this mineral. Emery is used as tip bits of cutting, drilling tools, etc. Artificial emery used for buffing and polishing is a graded combination of alumina and magnetite. (iv) Garnets: They are trisilicates of alumina, magnesia and ferrous oxide. The common garnet used as an abrasive is a complex of calcium–aluminium–iron silicates with the approximate formula: Ca3Al2(SiO4)3.Ca3Fe2(SiO4)3.Fe3Al2(SiO4)3 Hardness of garnets ranges from 6.0 to 7.5 on Moh’s scale. Garnets are too soft for grinding steel and iron, but when glued to paper or cloth, they are used for finishing hardwoods. They are also used for bearing pivots in watches, glass grinding, and polishing metals. (v) Quartz: Quartz is composed of silica (SiO2), which is impure and grey in colour. Hydrated form of quartz is the abrasive flint that is used on good quality sand paper. It is almost as hard as garnet; it is used for grinding, and other sharpening stones are cut from sandstone. This rock consists of quartz particles cemented together with feldspars, clays, carbonates and other minerals. It is used for grinding floor, pigments, ores, etc.

7.9.3 Artificial Abrasives Artificial abrasives are as follows: (i) Carborundum or Silicon Carbide (SiC): Carborundum is a bluish black crystallised artificial mineral with hardness between corundum and diamond. It is made by subjecting a mixture of silica and carbon (coke or coal) to high temperature (1650°C–2200°C) in an electric furnace. SiO2 + 3C → SiC + 2CO ↑ It is very hard, and in Moh’s scale, the hardness is about 9.3 and chemically inactive and can withstand the action of high temperature without damage, but it is not tough and is somewhat brittle. It is mainly used in cutting-wheels, abrasive papers and cloths. It is extensively used for grinding of materials of low tensile strength like cast iron, brass, bronze, porcelain, marble, finishing of leather, glass and optical grinding of lenses.

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(ii) Alundum (Al2O3): It is prepared by subjecting a mixture of calcined bauxite, coke and iron to high temperature, that is, about 4000°C in an electric arc furnace. The iron, titanium and silica impurities settle down at the bottom of the furnace, and after solidification, the hard crystalline alumina is separated, crushed and grounded. It is sold under zero number of trade names including Aloxite. Alundum or artificial corundum is not as hard as carborundum but is also less brittle and tougher. It is, therefore, used in preference to carborundum for grinding hard steels and other materials of high tensile strength or abrasive papers and cloths; it is used for finishing wood works. (iii) Boron carbide or Norbide (B4C): It is inert and one of the hardest artificial abrasives having hardness about 9 in Moh’s scale. It is made by heating boron oxide with coke in an electric furnace to approximately 2700°C. 2B2O3 + 7C → B4C + 6CO ↑ It is used on hard materials for making grinding dies and for cutting and sharpening hard, high-speed tools.

7.10 Review Questions 7.10.1 Fill in the Blanks 1.

are good conductors. [Ans.: Metals]

2. At absolute zero temperature, solid semiconductors act as perfect [Ans.: insulators] 3.

are important semiconductors. [Ans.: Si and Ge]

4. The materials which are used to prevent the loss of electricity through certain parts in an electrical systems are [Ans.: dielectrics] 5. Semiconductors that have defects due to impurity are [Ans.: extrinsic] 6. The addition of Group-15 elements to Group-14 elements gives [Ans.: n-Type] 7. At absolute temperature, the [Ans.: electrical resistance]

semiconductors. semiconductors.

of metal becomes zero.

8. The phenomena of metals that become perfect conductors with zero resistivity at absolute temperature is known as . [Ans.: super conductors] 9.

is the material possessing adhesive and cohesive properties and capable of bonding material. [Ans.: Cement]

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7.62 Engineering Chemistry 10. Natural cement is made by calcination of a naturally occurring [Ans.: argillaceous]

limestone.

11.

, an English mason, invented the cement. [Ans.: Aspdin Joseph]

12.

cement is the oldest cement invented by Romans. [Ans.: Puzzolana]

13.

cement resembled in colour and hardness to Portland stone. [Ans.: Portland]

14. The formula of gypsum is [Ans.: CaSO4 ⋅ 2H2O] 15.

acts as a retarding agent for early setting of cement [Ans.: Gypsum]

16. The process of solidification of cement consists [Ans.: setting, hardening] 17. Calcarious material is rich in [Ans.: lime]

and

Al O 18. According to ISI specifications of cement, the ratio of 2 3 shall not be less than Fe2 O3 [Ans.: 0.66] 19. Plaster of Paris setting and hardening [Ans.: alums or alkali sulphur] 20.

can initiate as well as hasten crystallisation.

are ceramic materials that can withstand high temperatures. [Ans.: Refractories]

21. Al2O3 and SiO2 are [Ans.: Acidic] 22.

refractories.

are the examples of basic refractories. [Ans.: CaO, MgO]

23. Introducing the lubricant to reduce frictional resistance between the moving or sliding surfaces is known as . [Ans.: lubrication] 24. Fluid film lubrication mechanism is also known as [Ans.: Thick-film or Hydrodynamic] 25.

oils are considered to be satisfactory lubricants for fluid film lubrication. [Ans.: Hydrocarbon]

26. Improving the properties of petroleum by incorporating specific additives is called [Ans.: blended oils]

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27. Calcium-based greases are emulsions of petroleum oils with [Ans.: calcium soaps] 28. The most usual solid lubricants are [Ans.: graphite and molybdenum disulphide]

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29. When graphite is dispersed in oil, it is called V. [Ans.: oil dag] 30.

is the property of a liquid or fluid by virtue of which it offers resistance to its over flow. [Ans.: viscosity]

31. Primary explosives are also known as [Ans.: detonators] 32. Propellants are known as [Ans.: low explosives]

. .

33. A mixture of 75% potassium nitrate, 15% charcoal and 10% sulphur is [Ans.: gun-powder] 34. Mixture of TNT and ammonium nitrate is [Ans.: amatol]

35. Principal ingredient in dynamites are [Ans.: nitroglycerine] 36.

is a thin, water-proof canvas length tube containing gunpowder or (TNT). [Ans.: A fuse]

37. A good propellant should have [Ans.: high specific impulse] 38.

is the energy liberated per unit mass of the explosives. [Ans.: Explosive strength]

39.

explosives have higher energy content than the primary explosives. [Ans.: High]

40.

is an extremely powerful, sensitive and standard military explosive. [Ans.: PETN]

41. TEM means . [Ans.: transmission electron microscopy] 42. SEM means . [Ans.: scanning electron microscopy] 43. XRD means [Ans.: X-ray diffraction]

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7.64 Engineering Chemistry 44.

crystals are highly anisotropic fluids that exists between the boundaries of the solid and liquid phase. [Ans.: Liquid]

45.

crystals move in three directions and can rotate about one axis. [Ans.: Nematic]

46.

crystal structure is optically active. [Ans.: Cholesteric]

47. Lyotropic crystal has [Ans.: amphiphile]

and water.

48.

mesophases are important in soaps, gels and colloids. [Ans.: Lyotropic]

49.

are the natural abrasives. [Ans.: Diamond, corundum, emery and quartz]

50. Carborundum Norbide are [Ans.: Artificial]

abrasives.

51. Hardness of an abrasive is measured roughly on [Ans.: Moh’s, Vicker’s]

scales.

7.10.2 Multiple-choice questions 1. Which of the following compounds act as super conductors? (a) YBa2Cu3O7 (b) Bi2Ca2Sr2Cu3O10 (c) Tl2Ca2Ba2Cu3O10 (d) All the above [Ans.: d] 2. Doping of Germanium with Group-13 elements gives (a) n-type semiconductors (b) p-type semiconductor (c) Super conductor (d) Conductor [Ans.: b] 3. Hydration of cement produces small volume changes known as (a) Dehydration (b) Soundness (c) Calcination (d) Hardening [Ans.: b] 4. The formula for plaster of Paris is (a) CaSO4 ⋅ H2O (c) CaSO4 ⋅ 2H2O [Ans.: c]

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(b) CaSO4 ⋅ 3H2O (d) CaSO4

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5. The formula for tricalcium alumina ferrite (a) 4CaO ⋅ Al2O3 ⋅ Fe2O3 (c) CaO ⋅ Al2O3 [Ans.: a]

(b) 4CaO ⋅ Fe2O3 (d) Al

6. Which of the following is responsible for high ultimate strength in cement? (a) C3S (b) C2S (c) C3A (d) C4AF [Ans.: a] 7. The major component of Portland cement is (a) MgO (b) C3A (c) CaO (d) C2A [Ans.: c] 8. Tobermonite gel is chemically (a) Hydrated tricalcium aluminate (c) Hydrated dicalcium ferrite [Ans.: b]

(b) Hydrated tricalcium silicate (d) Hydrated dicalcium silicate

9. The argillaceous material is rich in (a) Lime (c) Stone [Ans.: b]

(b) Silica (d) Ferriferrite

10. According to ISI specification, insoluble residue should not exceed (a) 5% (b) 3% (c) 10% (d) 2% [Ans.: d] 11. Any material that can withstand high temperatures without either breaking or suffering a deformation in shape is called (a) Dielectric (b) Thermal insulator (c) Refractory (d) Insulator [Ans.: c] 12. In acidic environment, refractory should not preferably be (a) Acidic (b) Basic (c) Neutral (d) None of these [Ans.: b] 13. Alumina is an example of (a) Acidic refractory (c) Neutral refractory [Ans.: a]

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(b) Basic refractory (d) None of these

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7.66 Engineering Chemistry 14. High resistance to spalling is shown by (a) Magnesia refractory (c) Alumina refractory [Ans.: c]

(b) Dolomite refractory (d) Lime refractory

15. Breaking, cracking or peeling off a refractory material under high temperature, is called (a) Thermal expansion (b) Spalling (c) Fusion (d) Chemical cracking [Ans.: b] 16. A refractory material, generally obtained from bauxite (a) Fire clay (c) Chromite [Ans.: d]

(b) Dolomite (d) Alumina

17. Which of the following is a neutral refractory? Dolomite (b) Graphite (c) Silica (d) Magnesia [Ans.: b] 18. Porosity in a refractory brick, generally, increases (a) Density (b) Resistance to spalling (c) Strength (d) Melting point [Ans.: b] 19. A good refractory material must (a) Be chemically inactive in use (b) Possess low softening temperature (c) Undergo spalling (d) Possess high thermal expansion [Ans.: a] 20. Spalling of refractory material can be reduced by (a) Rapid changes in temperature (b) Using porous refractory material (c) Using high coefficient of expansion refractory material (d) Using good thermal conductivity refractory material [Ans.: b] 21. Good thermal conductivity of a refractory material is desirable if it is to be used in the construction of walls of a (a) Blast furnace (b) Muffle furnace (c) Reverberatory furnace (d) All the above [Ans.: b] 22. The porosity of a refractory brick increases its (a) Strength (b) Abrasion resistance (c) Corrosion resistance (d) Resistance to thermal spalling [Ans.: d]

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23. Which one of the following refractories cannot be used in oxidising conditions? (a) Dolomite bricks (b) Magnesite bricks (c) Carbon bricks (d) Silica bricks [Ans.: c] 24. Which one of the following refractories is neutral in character? (a) Dolomite bricks (b) Silica bricks (c) Chromite bricks (d) Fire clay bricks [Ans.: c] 25. Which one of the following refractories is used in nuclear engineering as modulator? (a) Chromite bricks (b) Carborundum (c) Beryllia bricks (d) Fire clay bricks [Ans.: c] 26. An example of acid refractory is (a) Chromite (c) Silica [Ans.: c]

(b) Dolomite (d) Magnesite

27. Silica bricks belong to (a) Acidic refractories (c) Neutral refractories [Ans.: a]

(b) Basic refractories (d) None of these

28. High resistance to spalling is shown by (a) Magnesia refractory (c) Alumina refractory [Ans.: c]

(b) Dolomite refractory (d) Lime refractory

29. How does the porosity of refractory brick affect its mechanical strength? (a) Increases (b) Decreases (c) Slightly change (d) No change [Ans.: b] 30. Refractoriness is measured by using (a) Refractoriness under load (RUL) test (c) Conductivity test [Ans.: b]

(b) Seger cone test (d) Spalling test

31. Refractoriness under load is determined by (a) RUL test (b) Seger cone test (c) Conductivity test (d) None of these [Ans.: a] 32. A refractory should be (a) Chemically active (c) Chemically unstable [Ans.: b]

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(b) Chemically inactive (d) None of these

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7.68 Engineering Chemistry 33. Formula for porosity D −W W −D (a) P = × 100 (b) P = × 100 W−A A −W D −W W −D (c) P = × 100 (d) P = × 100 A −W W−A [Ans.: d]

W −D × 100, W indicates W−A (a) Weight of saturated specimen (b) Weight of dry specimen (c) Weight of dry specimen in water (d) Weight of saturated specimen in water [Ans.: a]

34. From the formula P =

35. Silica bricks are used in (a) Roofs of open hearth furnaces (c) Roofs of electric furnaces [Ans.: d]

(b) Coke oven walls (d) All the above

36. Refractory used in the linings of port land cement rotary kilns are (a) Silica bricks (b) Fire clay bricks (c) High-alumina bricks (d) Magnesite bricks [Ans.: c] 37. Magnesite bricks are used in (a) Acidic refractories (c) Neutral refractories [Ans.: b]

(b) Basic refractories (d) None of these

38. Refractories used in the construction of electrodes are (a) Magnesite (b) Alumina (c) Graphite (d) Chromite [Ans.: c] 39. Refractories used to separate acidic and basic refractory lining are (a) Chromite bricks (b) Alumina bricks (c) Graphite (d) All the above [Ans.: a] 40. Refractories used in high frequency electric furnaces are (a) Eryllia bricks (b) Zirconia bricks (c) Carborundum (d) None of these [Ans.: b] 41. Higher the porosity of refractory, (a) Higher the thermal conductivity (c) Lesser is its thermal conductivity [Ans.: a]

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(b) Higher the refractoriness (d) None of these

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42. The cause of thermal spalling is (a) Porosity (c) Maintaining low temperature [Ans.: b]

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(b) Rapid change of furnace temperature (d) None of these

43. Which are the most widely used refractories? (a) Fire clay bricks (b) Alumina bricks (c) Chromite bricks (d) None of these [Ans.: a] 44. High alumina bricks contain about 50–80% Al2O3 and 40–45% (a) MgO (b) CaO (c) SiO2 (d) Carbon [Ans.: c] 45. A refractory lining in a blast furnace should possess (a) Low thermal conductivity (b) High thermal conductivity (c) Medium thermal conductivity (d) None of these [Ans.: a] 46. A lubricant is used with the objective of (a) Increasing frictional heat (b) Increasing resistance (c) Decreasing frictional resistance (d) Providing direct contact between rubbing surfaces [Ans.: c] 47. A lubricant should possess high (a) Volatility (c) Oiliness [Ans.: c]

(b) Acidity (d) None of these

48. A lubricant is primarily used to prevent (a) Corrosion of metals (c) Oxidation of metal [Ans.: b]

(b) Wearing out of rubbing metallic surfaces (d) Reduction of metals

49. A suitable lubricant for watches is (a) Grease (c) Hazel-nut oil [Ans.: c]

(b) Graphite (d) Palm oil

50. A good lubricant should have (a) Low viscosity index (c) Low fire point [Ans.: b]

(b) High viscosity index (d) High

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7.70 Engineering Chemistry 51. The lubricant used in a machine working at low temperature should possess (a) High pour point (b) Low flash point (c) High cloud point (d) Low pour point [Ans.: d] 52. The capacity of an oil to stick on to the surfaces of machine parts under conditions of heavy load is called (a) Volatility (b) Oiliness (c) Acid value (d) Flash point [Ans.: b] 53. Oiliness is the least in case of (a) Greases (c) Animal oils [Ans.: b]

(b) Mineral oils (d) Palm oil

54. In case of liquid lubricants, generally (a) Flash point is higher than the fire point (b) Fire point is higher than the flash point (c) Fire point is lower than the flash point (d) Flash and fire points are identical [Ans.: b] 55. When the resistance to movement of sliding parts is only due to the internal resistance between the lubricant itself, the lubrication is called (a) Fluid film (b) Boundary (c) Thin film (d) Extreme pressure [Ans.: a] 56. Mineral oils are (a) Very costly (c) Unstable [Ans.: b]

(b) Poor in oiliness (d) Easily oxidised

57. Animal and vegetable oils are (a) Very cheap (c) Not thickened in use [Ans.: d]

(b) Not oxidised easily (d) Good in oiliness

58. Greases are not used to lubricate (a) Rail axle boxes (c) Gears [Ans.: d]

(b) Bearings working at high temperatures (d) Delicate instruments

59. Machines operating under high temperatures and loads are best lubricated by (a) Mineral oils (b) Solid lubricants (c) Greases (d) Animal oils [Ans.: b]

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60. When graphite is dispersed in oil, it is called (a) Grease (b) Aqua dag (c) Oil dag (d) Blended oil [Ans.: c] 61. Pick the odd one out. (a) Viscosity (c) Pour-point [Ans.: d]

(b) Carbon residue (d) RUL test

62. The single most important property of lubricating oil is its (a) Fire point (b) Cloud point (c) Oiliness (d) Viscosity index [Ans.: d] 63. What is the type of lubrication involved in delicate machines like watches? (a) Fluid-film (b) Thin film (c) Boundary (d) Extreme pressure [Ans.: a] 64. What are the type of oils suitable for thick film lubrication? (a) Animal oils (b) Vegetable oils (c) Blended oils (d) Hydro carbon oils [Ans.: d] 65. One of the important properties of greases is (a) Consistency (b) Oiliness (c) Thermal stability (d) All the above [Ans.: a] 66. The quality of grease will be measured by using (a) Refractometer (b) Caliometer (c) Penetrometer (d) Vaporimeter [Ans.: c] 67. Precipitation number indicates (a) Ash content (c) Moisture [Ans.: b]

(b) Asphalt (d) None of these

68. The aromatic content in the lubricant is determined by (a) Saponification number (b) Aniline point (c) Precipitation number (d) Neutralisation number [Ans.: b] 69. The estimation of carbon residue is generally carried out by (a) Grease penetrometer (b) Conrodson’s apparatus (c) Vapourimeter (d) None of these [Ans.: b]

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7.72 Engineering Chemistry 70. A good lubricant should deposit (a) More amount of carbon (c) more amount of ash [Ans.: b]

(b) Least amount of carbon (d) None of these

71. The oiliness of a lubricant will be increased by the addition of (a) Mineral (b) Vegetable or animal oil (c) grease (d) Solid lubricants [Ans.: b] 72. The diameter of the jet in Redwood No. 1 viscometer is (a) 1.42 mm (b) 1.62 mm (c) 3.4 mm (d) 3.6 mm [Ans.: b] 73. The jet borelength in Redwood No. 2 viscometer is (a) 5 mm (b) 10 mm (c) 15 mm (d) 20 mm [Ans.: b] 74. The formula used to measure the viscosity index is L −V U −L (a) VI = ×100 (b) VI = ×100 L −U V −L L −U L −V (c) VI = ×100 (d) VI = ×100 L−H H −L [Ans.: c] 75. Gulf oils are called (a) L–Oils (c) L and H Oils [Ans.: a]

(b) H – Oils (d) None of these

76. The unit of viscosity is (a) Calorie (c) Poise [Ans.: c]

(b) Eta (d) Seconds

77. The force per unit area F = is (a) h/d v (b) h ∝ v/ d (c)h/d/v (d) h ∝ d/v [Ans.: b] 78. The amount of oil collected during viscosity determination using redwood viscosity is (a) 50 ml (b) 60 ml (c) 200 ml (d) 400 ml [Ans.: a]

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79. Absolute viscosity will be calculated by using the formula (a) Kinematic viscosity × density (b) Commercial viscosity × density (c) Commercial viscosity/density (d) None of these [Ans.: a] 80. Abel’s apparatus is used to determine (a) Carbon residue (c) Viscosity [Ans.: d]

(b) Consistency (d) Flash point

81. Graphite and molebdenim disulphide are (a) Synthetic lubricants (c) Solid lubricants [Ans.: c]

(b) Antioscidants (d) Additives

82. The usual coefficient of friction between solid lubricants is between (a) 0.001 and 0.01 (b) 0.005 and 0.05 (c) 0.005 and 0.007 (d) 0.005 and 0.01 [Ans.: d] 83. Hexanol is a (a) Viscosity index improver (c) Corrosion prevention [Ans.: a]

(b) Thickness (d) Oiliness carrier

84. Lubricants also act as (a) healing agent (c) Cooling medium [Ans.: d]

(b) Corrosion preventer (d) All the above

85. The coefficient of friction in fluid film lubrication is (a) 0.01 to 0.03 (b) 0.01 to 0.3 (c) 0.01 to 0.05 (d) 0.001 to 0.03 [Ans.: d] 86. Tropex is a mixture of (a) 40% RDX, 40% TNT and 20% Al powder (b) TNT and PETN (c) 70% Tetryl and 30% TNT (d) None [Ans.: a] 87. Ammonium nitrate is a (a) Low single compound explosive (b) High single compound explosive (a) and (b) (d) None of these [Ans.: b]

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7.74 Engineering Chemistry 88. Lead azide is a (a) High explosive (c) Primary explosive [Ans.: c]

(b) Low explosive (d) Propellant

89. Gun powder is (a) Low explosive (c) Blasting fuse [Ans.: a]

(b) High explosive (d) All the above

90. A propellant used in rocket engine is (a) Fuel (c) None of these [Ans.: d]

(b) oxidant (d) (a) and (b)

91. The term nano means (a) One billionth of a kilometer (c) One billionth of an inch [Ans.: b]

(b) One billionth of a meter (d) One billionth of a millimeter

92. Who is the father of nanoscience? (a) Rutherford (c) Newton [Ans.: b]

(b) Richard Feynmen (d) Curie

93. Which of the following nanomaterials act as sensors of gases like NO2 and NH3 on the basis of increase in electrical conductivity? (a) Carbon nanotubes (b) Thin film (c) ZnO (d) Palladium [Ans.: a] 94. The nanotubes of MoS2 and CoS2 are used as (a) semiconductors (b) Insulators (c) Storage device (d) Solid lubricants [Ans.: d] 95. Which of the following nanowires shows photoluminescence? (a) Zinc oxide (b) Palladium (c) Silicone (d) MoS2 [Ans.: c] 96. Which nanomaterial effectively catalyses hydrogenation of oil? (a) Rane CuO (b) Rane pd (c) Rane Ni (d) Rane ZnO [Ans.: c] 97. Nanowires and nanotubes are (a) One dimensional (c) Three dimensional [Ans.: b]

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in nanoscale. (b) Two dimensional (d) None of these

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type of nanomaterials has a three-dimensional structure. (a) Thin film (b) Nanowires (c) Quantum dots (d) All the above [Ans.: c]

99. According to molecular arrangements, nanomaterials are broadly divided into (a) Two (b) Three (c) Four (d) Many [Ans.: b] 100. Thermotropic liquid crystals are (a) nantiotropic (c) (a) and (b) [Ans.: c]

types.

(b) Monotropic (d) None of these

101. Crystals which can change into the liquid crystal state from either lowering temperature of a liquid of raising of the temperature of solid are called (a) Nantiotropic (b) Monotropic (c) Nematic (d) All the above [Ans.: a] 102. Which mesophase is a nematic type of liquid crystal and optically active? (a) Meiotic (b) Cholesteric (c) a and b (d) None of these [Ans.: b] 103. The source of energy in nuclear fuel is due to (a) Chemical reaction (b) Nuclear reaction (c) Nuclear fission (d) Nuclear fusion [Ans.: c] 104. Hardness of an abrasive can be measured with (a) Rictor scale (b) pH scale (c) Moh’s scale (d) None of these [Ans.: c] 105. Diamond is (a) Natural abrasive (c) Inactive to acids, alkalis [Ans.: d]

(b) Crystallised carbon (d) All of these

106. The following are natural abrasives (a) Diamond (c) Garnet [Ans.: d]

(b) Emery (d) All of these

107. The following are artificial abrasives (a) Carborundum (c) Boron carbide [Ans.: d]

(b) Alundum (d) All of these

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7.10.3 Short Answer Questions 1. Define semiconductors. Ans.: Semiconductors are those solids which are perfect insulators at absolute zero, but conduct electric current at room temperature and lies between that of a good conductor and insulator. 2. What is meant by doping? Ans.: The addition of some impurities to pure semiconductors to enhance conductivity is known as doping. 3. Explain intrinsic semiconductors and conduction. Ans.: At normal temperature and pressure, pure silicon and germanium are insulators, because no electron is free and all the electrons are fixed in covalent bonds. However, at high temperature, covalent bonds are broken and the electrons so released become free to move in the crystal and thus conduct electric current. This type of conduction is known as intrinsic conduction and the conductors are intrinsic semiconductors. It will happen in the crystal without adding any external substance. 4. Classify extrinsic semiconductors with examples. Ans.: n-type semiconductors: Upon addition of trace amount of group 15 elements like phosphorous and arsenic to pure germanium, n-type semiconductors are formed. Extra electrons come from the impurity and conduct the electricity. P-type semiconductors: Upon addition of trace amount of group 13 elements like boron to pure germanium, p-type semiconductors are formed. Formed positive holes by doping are responsible for conduction. 5. Define superconductors and give examples. Ans.: Materials which show zero electrical resistance at absolute zero temperature are super conductors and the state is superconductivity. Ex: YBe2Cu3O7, Bi2Ca2Sr2Cu3O10. 6. Who invented the magnet and how was it named? Ans.: More than 2,000 years ago, the ancient Greeks first discovered a mineral that attracts things made of iron. This mineral was found in Magnesia, a part of Turkey; hence, it was named magnet. 7. Give the classifications of magnetic materials. Ans.: (i) Diamagnetic (ii) Paramagnetic (iii) Ferromagnetic (iv) Ferrimagnetic (v) Anti-ferromagnetic materials. 8. Give examples of diamagnetic materials. Ans.: Copper, silver, gold, etc., are the examples of diamagnetic materials. 9. Define hysteresis. Ans.: Retaining of magnetic properties after the removal of external magnetic field is known as hysteresis.

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10. What is meant by magnetic domain? Ans.: In ferro and ferromagnetic materials, below Curie temperature, a large number of atom moments is aligned parallels as a small volume region. These are known as magnetic domains. 11. Give the broad classifications of cement. Ans.: Cement is broadly classified into natural, puzzolana, slag and Portland cement. 12. Who invented puzzolana cement and how it can be prepared? Ans.: Puzzolana cement is the oldest cement invented by Romans. This is prepared by mixing of natural puzzolana and slaked lime. 13. What is known as the magic powder and what are its main ingredients? Ans.: Portland cement is also known as magic powder. It primarily consists of lime, silica, alumina and iron. 14. What is the function of iron oxide in cement? Ans.: Iron oxide provides colour, strength and hardness to the cement. 15. Give the equation for final setting and hardening of the cement. Ans.: 2C3S + 6H2O → C3S2 ⋅ 3H2O + 3Ca(OH)2 + 500 kj/kg Tricalcium silicate calcium hydroxide

Tobermonitegel

crystalline

16. What are the theories that explain the hardening of the cement? Ans.: Colloidal theory by Michaels and crystalline theory by Le Chatelier explain the hardening of the cement. 17. What are the chemical constituents present in Portland cement? Ans.: Tricalcium silicate (3CaO SiO2), dicalcium silicate (2CaO SiO2), tricalcium aluminate (3CaOAl2O3) and tetracalcium alumino ferrite (4CaOAl2O3Fe2O3). 18. What is the main function of gypsum in cement? Ans.: The presence of gypsum in the cement helps to retard the speed of the initial set, due to the formation of insoluble calcium sulphoaluminate. This does not show tendency to rapid hydration. C3A + Gypsum → Tricalcium sulphoaluminate 19. Define a refractory. Ans.: Refractories are ceramic materials that can withstand high temperatures as well as abrasive and corrosive actions of molten metals, slags and gases without deformation. 20. Explain any two important properties of a refractory. Ans.: Refractoriness: Refractoriness is the ability of a material to withstand heat without appreciable deformation. Dimension stability: It is the resistance of a material to any volume changes, which may occur on its exposure to high temperature and load. 21. Give porosity equation. Ans.: P =

W −D ×100 W−A

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7.78 Engineering Chemistry where W = weight of saturated specimen D = weight of dry specimen A = weight of saturated specimen submerged in water A good refractory should have low porosity. 22. Give examples of all types of refractories. Ans.: Acidic refractories: Alumina, silica Basic refractories: Magnesite, dolomite Neutral refractories: Graphite, zirconia 23. What is meant by permeability? Ans.: Permeability is a measure of rate of diffusion of gases, liquids and molten solids through a refractory. It depends upon the size and number of connected pores. Permeability will increase with temperature. 24. What are the uses of magnesite bricks? Ans.: Magnesite bricks are used in open hearth furnaces, reverberatory furnaces, rotary kilns and refining furnaces. 25. What is the test used to determine the refractoriness of the refractory? Ans.: Seger cone test. 26. Define thermal spalling. Ans.: Breaking, cracking, peeling off or fracturing of a refractory brick under high temperature is known as thermal spalling. 27. Define lubricants. Ans.: The process of reducing frictional resistance between moving or sliding surfaces by the introduction of lubricants in between them is called lubrication. 28. What are the main mechanisms of lubrication? Ans.: (i) Fluid film or thick film lubrication (ii) Boundary or thin film lubrication (iii) Extreme pressure lubrication 29. What are the main types of lubricants? Ans.: On the basis of physical state, lubricants are classified into liquid lubricants, greases or semisolid lubricants and solid lubricants. 30. Give some examples of greases. Ans.: Calcium-based greases, sodium-based greases, lithium-based greases and axle greases. 31. Why does graphite act as a good lubricant? Ans.: Due to its unique layer structure, graphite acts as a good lubricant. 32. Explain oildag and aquadag. Ans.: In the presence of tannin, the dispersion of graphite in oil is called oildag and dispersion of graphite in water is called aquadag. 33. What are the units of absolute viscosity? Ans.: Eta (h) or poise or centipoise are units for absolute viscosity.

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34. Give the equation for viscosity index. Ans.: Viscosity index =

VL × VX × 100 VL − VH

35. Name the different types of viscometers. Ans.: Redwood viscometers, Engler’s viscometers, Saybolt viscometers, U-tube viscometers, etc. 36. How can we measure the mechanical stability of a lubricant? Ans.: Four-balls extreme pressure is the one of the important mechanical tests to judge the mechanical stability of a lubricant under load. 37. Give some important properties of explosives. Ans.: Explosive strength, velocity of detonation, sensitivity, brisance, etc., are some important properties of explosives. 38. Give some examples for detonators. Ans.: Lead azide, mercury fulminate, tetracene and diazodinitro phenol. 39. What is the main difference between detonators and propellants? Ans.: Detonators are highly sensitive and can explode with slight shock or fire, whereas propellants burn simply but do not explode. 40. How are high explosives classified? Ans.: Depending on the components present on the explosives, they are classified into single compound explosives, binary explosives, plastic explosives and dynamites. 41. What is the main component present in the binary explosives? Give some examples. Ans.: T NT is the main component present in binary explosives. Examples are amatol, pentolite, tetrytol, tropex, etc. 42. Define blasting fuses. Ans.: A fuse is a thin water-proof canvas length of tube containing gunpowder, arranged to burn at a given speed for setting of charges of explosives. 43. Define rocket propellants and give their classification. Ans.: A rocket propellant is high oxygen containing fuel and oxidant, whose combustion takes place in a definite and controlled manner with the evolution of a huge volume of a gas. Propellants are mainly classified into solid and liquid propellants. 44. What are the advantages of liquid propellants? Ans.: Liquid propellants possess many advantages over solid propellants because they are versatile and the engine using them can be checked and calibrated more easily. The engine using the liquid propellant is quite delicate and cannot withstand any rough handling. 45. Define a nanometre. Ans.: Nanometre is the one-billionth of a metre, 1 nm = 1/1000000000 metre = 1 × 10 –9 m. 46. Classify nanomaterials according to molecular arrangements. Ans.: According to atoms/molecular arrangements, nanomaterials are broadly classified into three types.

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7.80 Engineering Chemistry Materials which have one dimension in the nanoscale. For example,, surface coatings, thin films, etc. Materials which have two dimensions in the nanoscale. For example, nanowires, nanotubes, etc. Materials which have three dimensions in nanoscale or quantum dots. For example,, fullerenes. 47. What are the important properties of nanomaterials? Ans.: A significant increase in surface-area-to-volume ratio at the nanoscale gives rise to novel and enhanced magnetic, mechanical, electronic, catalytic, conducting and optical properties. 48. Give some examples of semiconducting and superconducting nanomaterials. Ans.: Semiconductors: Carbon nanotubes, nanowires, MoS2, etc. Superconductors at high temperature: NbS2. 49. What are the most important applications of carbon nanotubes? Ans.: Well-defined geometry, exceptional mechanical properties and extraordinary electrical characteristics of carbon nanotubes are used in nanoelectric circuits, nanoelectro mechanical systems, nanorobotics, nanobiosensors, etc. 50. Give the interconversions of a nanometer. Ans.: One nanometre is the one-billionth of a meter. 1 nm = 10 – 9 m = 10 – 9 yards approximately. 51. Give the different synthetic routes of nanomaterials. Ans.: The vapour–liquid–solid growth, solution–liquid, solid growth, template-mediated growth, electron beam lithography, reverse micellar route, etc. 52. How can we characterise nanomaterials? Ans.: Nanomaterials are characterised by X-ray diffractions (XRD), transmission electron microscopy (TEM), scanning electron microscopy (SEM), thermal analysis like TGA/ DTA, Fourier transform infrared spectroscopy (FITR), etc. 53. Give the broad classification of nanomaterials. Ans.: (i) Materials which have one dimension in the nanoscale. For example,, surface coatings or thin films. (ii) Materials which have two dimension in the nanoscale. For example,, nanowires and nanotubes. (iii) Materials which have three dimension in the nanoscale. For example,, quantum dots. 54. Which property plays a vital role in nanomaterials? Ans.: A significance increase in surface area-to-volume ratio at the nanoscale gives rise to novel and enhanced magnetic, mechanical, electronic, catalytic, conducting, optical properties, etc. 55. Give a brief note about fullerenes and carbon nanotubes. Ans.: Fullerenes are a class of allotropes of carbon, which are basically graphene sheets rolled into tubes or spheres. They include carbon nanotubes because of their mechanical strength and electrical properties. 56. What are the important uses of carbon nanotubes? Ans.: Nanoelectronic circuits, nanoelectromechanics, nanorobotics, probes, grippers, nanobiosensors, etc., are important uses of carbon nanotubes.

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57. Give the broad classification of liquid crystals. Ans.: Liquid crystals are broadly classified into thermotropic liquid crystals and hydropic liquid crystals. 58. Name the three kinds of thermotropic liquid crystals. Ans.: The three types are nematic, cholesteric and smectic liquid crystals. 59. What is the main difference between nematic and cholesteric structure? Ans.: Cholesteric structure is a nematic type of liquid crystal but it is optically active. 60. Give examples of nematic liquid crystals. Ans.: P-methoxybenzylidene p1-N-butyl-aniline and P-n-Hexyl-P1-cyanobiphenyl. 61. Why are lyotropic crystals known as amphiphilic? Ans.: Lyotropic crystals are composed of both lyotropic and lyophobic parts. 62. Define liquid crystals. Why are they named liquid crystals? Ans.: Liquid crystals are highly anisotropic fluids that exists between the boundaries of the solid and conventional, isotropic liquid phase. Liquid crystals are made up of both liquid and crystal. The term “liquid” is used because the tendency to take the shape of the container and the word crystal is used because they still contain one- to two-dimensional arrays. 63. Name the different types of thermotropic liquid crystals. Ans.: Nematic, cholesteric and smectic structures are main types of thermotropic liquid crystals. 64. Define lyotropic liquid crystal and give an example. Ans.: Lyotropic liquid crystals are mixtures of two or more components that change phase with changes of concentration. For example,, sodium stearate, a -Lecithin. 65. Give some applications of liquid crystals. Ans.: Liquid crystals are widely used in research, medicine, displays, radiation sensors, thermometers, non-destructive testers, etc. 66. Give the importance of liquid crystals in medicine. Ans.: Liquid crystals are widely used in medicine as optical disks, full colour electronic slides, light modulators, etc., because most biological systems exhibit liquid crystal properties. 67. What is the reason for liquid crystals being used as thermometers? Ans.: The temperature-dependent variation in the colour of cholesteric liquid crystals has led to the use of substances in the measurement of temperature and gradients of temperature. A cholesteric substance or a mixture of cholesterics always exhibits the same colour and the same temperature. Hence, these are also used to locate veins, arteries, infections, tumours, etc., which are warmer than surrounding tissues. 68. Give commonly used natural and artificial abrasives in the order of increasing hardness. Ans.: Talc < Gypsum or NaCl < Calcite < Fluorite < Apatite < Feldspar < Quartz < Topaz or emery < Corundum < Diamond

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7.82 Engineering Chemistry 69. Give some common uses of abrasives. Ans.: (i) Housewives and farmers use abrasive stones to keep their kitchen knives and agricultural instruments sharp. (ii) Dentists use an abrasive powder when they clean teeth and to smooth down fillings. (iii) Abrasives play important roles in various industries such as in grinding of wood into paper pulp, cutting of stone into carved and surfaced structures, and sharpening of cutting-tools. 70. Give broad classification of abrasives. Ans.: Abrasives are broadly classified into natural and artificial abrasives. 71. Give brief note about garnet. Ans.: Garnets are trisilicates of alumina, magnesia and ferrous oxide. The common garnet used as an abrasive is a complex of calcium–aluminium–iron silicates with the approximate formula: Ca3Al2(SiO4)3.Ca3Fe2(SiO4)3.Fe3Al2(SiO4)3 Hardness of garnets ranges from 6.0 to 7.5 on Moh’s scale. Garnets are too soft for grinding steel and iron, but when glued to paper or cloth, they are used for finishing hardwoods. They are also used for bearing pivots in watches, glass grinding, and polishing metals.

7.10.4 Descriptive Questions Q.1 Define semiconductors and explain intrinsic and extrinsic semiconductors with example. Q.2 Write a note on n-type and p-type semi-conductors. Q.3 Give a brief note on the importance and application of semi-conductors. Q.4 Explain super conductors with examples. Q.5 Define magnetism and explain dia and paramagnetic materials. Q.6 Explain the classification of magnetic materials in detail. Q.7 Give a brief note on the important properties of magnetic materials. Q.8 Explain the applications of magnetic materials. Q.9 What is Portland cement? Explain the different ingredients of cement. Q.10 Give an account of the following: (a) Chemical composition of cement (b) Chemical constitution of Portland cement (c) Naming of Portland cement Q.11 Explain the setting and hardening of cement with suitable chemical reactions. Q.12 Write a brief account on the following: (a) Raw materials and the ingredients of cement (b) Function of gypsum in cement (c) Discuss merits and demerits of dry process and wet process. Q.13 Explain the analysis of cement.

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Q.14 Draw a labelled diagram of a rotary kiln used for the manufacture of Portland cement by wet process and discuss the various reactions taking place in the furnace. Q.15 What are the microscopic constituents or constitutional compounds present in Portland cement? How do they contribute towards the properties of the cement? Q.16 What do you mean by setting and hardening of cement? Discuss the various reactions involved with the help of equations. Q.17 “The properties of Portland cement depend upon the relative proportions of its constitutional compounds”. Justify the statement. Q.18 What are the different methods of manufacturing cement? Discuss their relative merits and demerits. Q.19 Write informative notes on the following: (a) Reactions taking place in the rotary kiln. (b) Constitutional compounds in cement and its derivatives. (c) Additives for cement. (d) Important properties of cement. (e) How is cement classified? (f) Define the soundness of cement. Q.20 (a) W hat is a pyrometric cone equivalent? How is it determined for a refractory? What is its significance? (b) Write a short note on the following: (i) Porosity (ii) Thermal conductivity (iii) Dimensional stability (iv) Strength Q.21 (a) Define refractories and what are the criteria of a good refractory? (b) Give the classification of refractories with suitable examples. Q.22 (a) W hat are refractories? How important are the properties of refractoriness under load and thermal conductivity for industrial applications? (b) Compare acidic and basic refractories with examples. Q.23 (a) How are refractories classified? Give one example for each class. (b) Write a note on the conditions leading to failure of a refractory material. Q.24 Discuss any four essential properties of a good refractory in detail. Q.25 What are the causes leading to failure of a refractory? Q.26 Write short note on the following: (a) Refractoriness (b) Refractoriness under load or strength (c) Dimensional stability (d) Thermal conductivity (e) Porosity

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7.84 Engineering Chemistry Q.27 State some important industrial applications of refractories. Q.28 Discuss the important properties of refractories which have a direct bearing on their industrial use. Q.29 Write informative notes on the following: (a) Fire clay refractories (b) Silica refractories (c) Magnesite refractories Q.30 (a) What are the raw materials for refractories? (b) What are the different steps in the manufacture of refractories? (c) What do you mean by super refractories? Q.31 Discuss the various physical and chemical factors which affect the industrial uses of refractories. Q.32 (a) Give the functions of lubricants. (b) Describe the mechanism of extreme pressure lubrication. (c) How is a viscous lubricant converted into grease? Q.33 Discuss the important properties of lubricating oils, which are useful for their evaluation. Q.34 (a) Distinguish between hydrodynamic lubrication and boundary lubrication. (b) Distinguish between hydrodynamic lubrication and extreme pressure lubrication. Q.35 Explain the following two theories for the mechanism of the lubricants: (a) Boundary lubrication (b) Extreme pressure lubrication Q.36 Write notes on the following: (a) Blended oils (b) Petroleum oils (c) Extreme pressure additives (d) Antioxidants Q.37 Explain how the following act as lubricants: (a) Graphite (b) Molybdenum disulphide Q.38 Write a note on lubricants with special reference to their classification, mode of action, examples and applications. Q.39 How do we select lubricants for the following? (a) Cutting tools (b) I.C. engines (c) Steam engines (d) Steam turbines (e) Gears Q.40 Explain the various mechanisms of lubrication in detail. Q.41 Define lubricants. Discuss the important properties of the lubricating oils.

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Q.42 Describe the various types of lubrication. Q.43 Define the flash and fire points. Q.44 (a) Describe thick-film lubrication. (b) Write a note on semi-solid lubricants. Q.45 Write short notes on the following properties of lubricants: (a) Pour point (b) Aniline point Q.46 Explain the hydrodynamic lubrication. Q.47 Explain the following properties of lubricants and discuss their significance: (a) Viscosity and viscosity index (b) Flash point (c) Aniline point (d) Saponification value Q.48 Distinguish between fluid film and boundary lubrication. Q.49 Lubricating oil has the same viscosity as standard naphthenic and paraffinic type oils at 210°F. Their viscosities at 100°F are 320 Saybolt universal second (SUS), 430 SUS and 260 SUS respectively. Find the viscosity index of the oil. Q.50 (a) What do you mean by viscosity index of lubricating oil? (b) Lubricating oil has a SUS of 58 seconds at 210°F and of 600 seconds at 100°F. The high viscosity index standard (i.e., Pennsylvanian) oil has Saybolt universal viscosity (SUV) of 58 seconds at 210°F and 400 seconds at 100°F. The low viscosity index standard (i.e., Gulf ) oil has a SUV of 58 seconds at 210°F and 800 seconds at 100°F. Calculate the viscosity index of oil. Q.51 Write an essay on solid lubricants with emphasis on their classification, mechanism of action, examples and applications. Q.52 How are semi-solid lubricants prepared? In what situations is a semi-solid lubricant preferred? Mention some important tests for evaluating semi-solid lubrications. Q.53 How are liquid lubricants classified? Discuss the various methods available for refining mineral oils. 54. What do you mean by blended or compounded oils? What are the various additives used to induce or improve the necessary properties of lubricating oil? Q.55 Discuss the use of lubricating emulsions. Q.56 Write informative notes on the following: (a) Cup greases (b) Fixed oils (c) Gas lubrication (d) Biodegradable lubricants (e) Neutralisation number (f) Extreme pressure lubrication (g) Oiliness

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7.86 Engineering Chemistry (h) Redwood viscometer (i) Cryogenic bearing lubricants ( j) Lubricants for nuclear reactor systems (k) Lubricants for food processing (l) Environmental and health factors in the use of lubricant Q.57 Justify the following statements: (a) Flash point determination by the closed-cup apparatus gives a lower value than that determined by an open-cup apparatus. (b) Closed-cup apparatus gives more reliable value than the open-cup apparatus for the determination of flash point. (c) The relative viscosity determined by Saybolt viscometer or Redwood viscometer can be converted into absolute kinematic viscosity by calculations. Q.58 Write short notes on saponification and iodine values. Q.59 Discuss the significance of viscosity in lubricating oil. How is it determined by Redwood viscometer? Q.60 (a) Write the structure of graphite. Based on this, suggest why this can be used as a solid lubricant. (b) Discuss the classification of lubricants. Q.61 Define lubricants. Discuss the classification of lubricant with suitable examples. Q.62 (a) Explain the following properties of lubricants and their significance: (i) Carbon residue (ii) Aniline point (b) Write an informative note on synthetic lubricants. Q.63 Define the term lubricants. Mention their important functions. Explain and discuss the significance of any two properties of lubricants. Q.64 Discuss lubrication, its mechanism and significance in brief. Explain viscosity index of lubricating oil. Q.65 (a) Define the terms lubrication and lubricants. What are the different types of lubricants? Discuss the basic principle of lubrication. (b) What are the chief functions of lubricants? (c) What are the different types of lubricants? (d) Discuss the classification of lubricants with an example. Q.66 (a) How are lubricating oils produced and refined? Which oil is a better lubricant? (b) Define dewaxing. What are the characteristic features of synthetic lubricating oils? Q.67 (a) Write the names of two semi-solid lubricants. (b) Give brief note on graphite and molybdenum disulphide. (c) Write a short note on the following: (i) Semi-solid lubricants (ii) Synthetic lubricants (iii) Solid lubricant and its advantage

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Q.68 (a) Write a short note on extreme-pressure lubrication? (b) W hat are the different synthetic lubricants used? How are they superior over petroleum lubricants? Q.69 (a) W hat are greases and under what situations are they employed? Discuss the composition and uses of the following: (i) Calcium-based greases (ii) Soda-based greases (iii) Axle greases (b) Write a note on extreme-pressure additives to mineral oil. Q.70 (a) Draw a neat and labelled diagram of the determination viscosity of lubricant by Redwood viscometer. (b) Write a short note on aniline point. Q.71 (a) How does viscosity determine the operating characteristics of the lubricants? (b) Suggest the suitable properties of lubricating oil used for steam engines and transformers. Q.72 (a) Cotton seed oil is used as dry oil in paints. Is it true or false? (b) Write a note on additive for lubricating oils. Q.73 (a) How are greases made? (b) What is a lubricant? Discuss the classification and its basic characteristics with examples. Q.74 (a) What are the characteristic features of synthetic lubricating oils? (b) Write an explanatory note on solid lubricants. Q.75 (a) Name any four solid lubricants. (b) Describe any four desirable properties of lubricant oil. (c) Write brief note on greases. Q.76 (a) Explain clearly the importance of the following in selecting lubricating oil for a particular use: (i) Viscosity (ii) Flash point (iii) Acidity (iv) Carbon residue (b) How is the viscosity of lubricating oil determined in the laboratory? Q.77 (a) How will you select a lubricant? (b) Explain the properties of lubricants such as viscosity and viscosity index. Q.78 (a) What is meant by lubricant? Explain the mechanism of lubrication. (b) Write a note on the following: (i) Lubricating greases (ii) Lubricant emulsions (c) Describe a method to manufacture lubricating oils. (d) Write a note on the lubricating action of greases. Q.79 (a) How are lubricants classified? (b) Define flash point. Describe any one method of determining flash point. (c) With the help of a neat diagram, explain the working of Redwood viscometer. (d) What is lubrication? Explain any one type of lubrication in detail.

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7.88 Engineering Chemistry Q.80 (a) Why are fatty oils no longer used as lubricants? (b) What do you mean by viscosity of a lubricating fluid? How does it change with temperature? How do viscosity and viscosity index influence the selection of lubricants for particular purposes? (c) W hat are flash point and fire point of a liquid lubricant? Are they directly related to the quality of lubricants? (d) Write a short account on solid lubricants. Q.81 (a) Under what situation greases are used? What are the main functions of soap in grease? (b) What do you understand by consistency and drop point of grease? Explain their importance. Q.82 (a) What is meant by “oiliness” of a lubricant? How can this be improved? (b) Explain boundary and fluid lubrications and mention the conditions therein. Q.83 (a) What are propellants and explains the characteristics of a good propellants? (b) Explain the properties of solid and liquid propellants. Q.84 (a) What are explosives? What are the basic requirement of chemical explosives? (b) Write short note on Dynamite. (c) Write short note on TNT. (d) Distinguish between primary and secondary explosives. Q.85 Write a short note on the following: (a) Classification of explosives (b) Plastic explosives (c) Rocket fuels (d) Propellants (e) Blasting fuses. Q.86 (a) What is detonation? (b) What are the requirements of a good propellants? (c) What are the requirements of a good explosives? Q.87 (a) What two factors for the selection of a propellant? (b) Explain Mono and bi propellants with examples. Q.88 Define nanomaterial. Explain some of the important properties. Q.89 Give a brief description about the applications of nanomaterials. Q.90 What are nanomaterials? Explain their characteristics. Q.91 Give broad classification of nanomaterial with an example. Q.92 Explain the following properties of nanomaterial: (a) Catalytic property (b) Mechanical property (c) Electrical property (d) Optical property Q.93 What are nanoparticles? Give brief description on their properties. Q.94 Give brief explanation about preparation of nanomaterials and their importance.

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Chemistry of Engineering Materials

7.89

Q.95 Give brief explanation about characterisation of nanomaterials. Q.96 Discuss about carbon nanotubes and their importance. Q.97 Define liquid crystal and explain the characterising of a liquid crystal phase. Q.98 Give a brief explanation about thermotropic liquid crystals with example. Q.99 Write short note on lyotropic liquid crystals. Q.100 Explain the important applications of liquid crystals. Q.101 Give brief note on the following: (a) Smectic structure (b) Enantiotropic crystal (c) Monotropic crystals. Q.102 Define abrasive and explain hardness of abrasive with Moh’s scale. Q.103 Give brief note on natural and artificial abrasives. Q.104 Write short note on the following: (a) Diamond (b) Carborundum (c) Alundum (d) Hardness of abrasive

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Phase Rule

8.1 Introduction In 1875, as a result of mathematical and thermodynamic studies, J. Willard Gibbs put forward a rule known as Phase Rule; without any exception, the rule is applicable to all heterogeneous system in equilibrium. By using Phase Rule, the effect of temperature, pressure and concentration can be predicted qualitatively on a heterogeneous system, which is in equilibrium by the Phase diagrams. It is assumed that the equilibrium is influenced only by temperature, pressure and concentration, but not influenced by gravity, electrical or magnetic forces or by surface action. The maximum number of degree of freedom is taken as three; mathematically, Gibbs’ Phase Rule may be stated as: F = C − P + 2 or F + P = C + 2 Where F = Number of degree of freedom C = Number of components P = Number of Phases

8.2 Explanation of the Terms involved in Phase Equilibria 8.2.1 Phase (P) A heterogeneous system consists of various homogenous parts in contact with each other by distinct boundaries; any part of a system which is homogeneous, physically distinct and mechanically separable from other parts of the system is a Phase. Essential Conditions for a Phase (i) It should be homogeneous or may be a homogeneous part of a heterogeneous system. (ii) It should be physically distinct and is separated from other parts of a system by well defined boundary surface. (iii) A dynamic equilibrium between different phases of this system should be present to exchange of chemical species.

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8.2 Engineering Chemistry Examples (i) A gas or a gaseous mixture is a single phase because there is no interface between one gas and another e.g. air which is a mixture of nitrogen, oxygen, carbon dioxide, water vapour, etc is composed of one phase only. (ii) Water exists in three forms-ice, water and vapour. It is a three phase system. solid (Ice) liquid (water) Gas (water vapour) (iii) Two or more completely miscible liquids present in a system constitute only one liquid phase as there will be no surface of separation between them when they are mixed. e.g.: water, alcohol, acetone etc. (iv) Two immiscible liquid forms two different phases. e.g.: water and ether, alcohol and ether etc. Both forms two different phase. (v) A heterogenous mixture of solid substances consists of as many phases as there are substances present in that system. e.g.: decomposition of CaCO3(s) CaCO3 (s) CaO(s) + CO2 (g) Here are three phases among those two phases are CaCO3(s) and CaO(s) and third phases is CO2(g) because all phases are separated by interface.

8.2.2 Components (C) The smallest number of independently variable constituents by which the composition of each phase present can be expressed directly or in the form of a chemical equation is the number of components (C) of a system at equilibrium. The number of components of a system may be or may not be the same as the actual number of substances or constituents present in the system; only those constituents of an equilibrium mixture, which can undergo independent variation, are known as components. (i) When no reaction takes place, the number of components in equal to the number of constituents. (ii) When a reaction occurs, at that time, then the minimum number of species allowing for a reaction to prepare one species is known as the number of components. Therefore, the number of components of a system C=S−R Where S = number of chemical species present in the system R = number of independent chemical reactions taking place among the chemical species. In case of Ionic Reaction The number of component is calculated as C = S − (R + 1) S = Total number of ionic species present in the system. R = number of relation between ionic species.

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Phase Rule

8.3

Examples (i) Water exists in three phases as Ice Water Vapour

(solid)

(liquid)

(gas)

Each phase can be represented by H2O. Thus, the number of components is one. (ii) Consider aqueous solution of sugar. Here, the composition of this solution is described by specifying the presence of sugar and water. Thus, the number of components is two. (iii) Consider the decomposition of Ammonium chloride as NH 4 Cl(s) NH 3 (g) + HCl(g) Actually this equilibrium exists as NH 4 Cl(s) NH 4 Cl(g) NH 3 (g) + HCl(g) (a) If the reaction is carried out in a closed vessel or vacuum then number of component is one because in the gaseous phase both HCl and NH3 are always present in equal amounts and represents NH4Cl(g). (b) If the excess of either NH3 or HCl is introduced, the composition of gaseous phase can no longer be the same and the composition of this phase can no longer be represented by NH4Cl alone, but one more component is required. Hence, it becomes two component system. (iv) Consider the thermal decomposition of CaCO3 as CaCO3 (s) CaO(s) + CO2 ( g ) There are S = 3 ﹛CaCO3(s), CaO(s), CO2(g)﹜ R=1 So, C = S − R =3−1=2 Hence, the number of component is two. Here, three different constituents form three different phases, but the composition of each phase can be expressed in terms of any two of the constituents. (a) If CaO and CO2 are chosen as the components, then Phase Composition CaCO3 → CaO + CO2 CaO → CaO + 0CO2 CO2 → 0CaO + CO2 (b) If CaCO3 and CO2 are chosen as the components, then Phase Composition CaCO3 → CaCO3 + 0CO2 CaO → CaCO3 − CO2 CO2 → 0CaCO3 + CO2 (c) If CaCO3 and CaO are chosen as the components, then Phase Composition CaCO3 → CaCO3 + 0CaO

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8.4 Engineering Chemistry CaO → 0CaCO3 + CaO CO2 → CaCO3 − CaO Thus, in all these cases the smallest number of constituents which can fix the composition of the phase present at equilibrium is two; hence, dissociation of CaCO3 by heat is a two component system. (v) Consider the system NaBr−KCl−H2O The number of components is calculated by formula C = S − (R + 1) Here number of species, S = 9 (NaBr, NaCl, KCl, KBr, H2O, Na+, Br−, K+, Cl−) (Dissociation of water is ignored) Number independent reactions (Relation) R=4 + − NaBr Na + Br + − KCl K + Cl

Na + + Cl − NaCl K + + Br − KBr ∴ Number of components C = S − (R + 1) = 9 − (4 + 1) =4 Hence, it is a four compound system. (vi) Consider the dilute solution of sulphuric acid in water + − H 2SO 4 + H 2 O H 3O + H SO 4 − + 2− H SO 4 H + SO 4 + − 2− Here, number of species, S = 5( H 2SO 4 , H 2 O, H 3O , HSO 4 , SO 4 ) Number of relation, R = 2 C = S − (R + 1) = 5 − (2 + 1) = 2 ∴ Hence, it is a two component system.

8.2.3 Degree of Freedom (F) Degree of freedom is defined as the number of intensive variables (temperature, pressure and concentration) that can be changed independently without disturbing the number of phases in equilibrium. These variables describe the state of the system. On the basis of degree of freedom of the system, systems are classified as nonvarient(F = 0), monovarient (F = 1), bivarient (F = 2) etc. Examples (i) Consider the example of pure gas, the number of degree of freedom is two (F = 2). This is because a pure gas satisfied the gas equation PV = RT. If the values of pressure (P) and

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Phase Rule

8.5

temperature (T) are fixed, then the volume (V) automatically gets fixed. In fact if any two variables (out of P, V & T) are specified. The third one gets specified by itself. Hence a system consisting of pure gas has two degree of freedom i.e. it is a bivarient system. (ii) Consider one component having two phase system. Water Vapour This system consists of two phases of one component. The vapour pressure of water is definite at a definite temperature independent on the concentration. It follows, therefore, that if the temperature is fixed, the vapour pressure is also fixed and vice-versa. So, we cannot alter both the variables without disturbing the equilibrium. Hence, we have to mention only one variables without disturbing the equilibrium i.e. either temperature or pressure. Thus the system has one degree of freedom (F = 1). System is called as monovarient or univarient (F = 1). (iii) Consider one component, three phases in equilibrium Ice Water Vapour In this system, there are three phases of one component (H2O). These three phases can co-exist in equilibrium only at particular temperature (0.0098 °C) and under one particular pressure (4.58 mm of Hg). Any variation of these factors will result into disappearance of one or more of the phases. Hence this system has zero degree of freedom (F = 0) i.e.nonvariant or invariant.

8.2.4 True and Metastable Equilibrium Under a given set of conditions, the same state of a system can be attained by approaching from either direction by any possible procedure; then, that system is said be in the state of true equilibrium. According to thermodynamics, true equilibrium is attained when the free energy content of the system is at a minimum for the given values of the variables. Example: At 1 atm pressure and 0 °C (273 K) ice and liquid water attain true equilibrium because it can be attained by either partial melting of ice or partial freezing of water. Ice Water Under a given set of conditions a state of a system can be attain by only one direction with careful changing of a system conditions; that system is said to be state of meta stable equilibrium. Example: At 271 K (−2 °C) or at even lower temperature, it is possible to cool water very slowly and carefully without appearance of ice; hence, water at −2 °C is said to be in a state of metastable equilibrium. The metastable equilibrium state of the system may be preserved by not subjecting it to a sudden shock, stirring or seeding by the solid phase. Solidification sets rapidly as soon as a crystal of ice is introduced and the temperature rises to 0 °C (273 K).

8.2.5 Eutectic Mixture and Eutectic Point Mixture of two or more components without reacting chemically in solution state and at a particular temperature having lowest freezing or melting point among all possible ratio of mixing of that component is known as eutectic mixture, and such a type of system is called the eutectic system and the corresponding lowest freezing or melting point of that eutectic mixture is called the eutectic point. For example in a eutectic mixture of lead (Pb) and silver(Ag) forms a eutectic point at 303 °C having the lowest melting point of eutectic system.

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8.6 Engineering Chemistry

8.2.6 Triple Point Triple point is a point at which three phases of a system co-exists in equilibrium. The degree of freedom of one component system is zero. At triple point, system is in variant i.e. degree of freedom is zero (F = 0). For example, consider a system of water having three phases Ice Liquid Vapour All these phase co-exist at a particular temperature (0.0098 °C) and a particular pressure (4.58 mm of Hg). At triple point F = 0 (invariant or non-variant) C = 1, P = 3, F=C−P+2=1−3+2=0 F = 0 (Invariant)

8.3 Phase Rule The phase rule gives relationship between the numbers of phases, components and degree of freedom of a heterogenous system. The Gibbs phase rule may be stated as follows: In a heterogenous system in equilibrium, the number of degree of freedom plus the number of phases is equal to the number of components plus two. Mathematically, F + P = C + 2 or F = C − P + 2 Where F = number of degree of freedom C = number of components P = number of phases Digit ‘2’ represents the temperature and pressure. It is assumed that the equilibrium is not influenced by gravity, electrical or magnetic forces or by surface action and is influenced only by temperature, pressure and concentration. The maximum number of degree of freedom is taken as three.

8.3.1 Assumptions for the Validation of Phase Rule (i) The system is in thermal and mechanical equilibrium; consequently, the pressure and temperature are the same in all the phases of system. (ii) Surface contribution as well contributions from any electric or magnetic field to the extensive properties of the system, are negligible. (iii) Inter phase surfaces are deformable and permeable to components.

8.3.2 Thermodynamic Derivation of the Phase Rule Let us consider a heterogenous system having C components (C1, C2, C3,…,Cc) distributed between P phases (P1, P2, P3,…,PP) as shown in Figure 8.1. As we know that the degree of freedom of a system is equal to the number of independent variables which must be fixed arbitrarily to define system completely.

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Phase Rule

8.7

Degree of freedom = Total number of variables − (Number of variables defined by the system because of its being in equilibrium) F=V−E

Components

Phases P1

Figure 8.1 A system showing C components distributed in P phases in equilibrium (i) Total number of variable can be calculated as follows: (a) Temperature: Same for all phase (one variable) (b) Pressure: Same for all phase (one variable) (c) Concentration: Suppose a phase contains two components (A and B). Then if molar concentration of one component is known, then that of other can be calculated by using formula x A + x B = 1. Similarly, if a phase contains three components, only two concentration terms should be known (xA + xB + xC = 1). Hence, in general if a phase contains C components, it can be defined completely by (C − 1) concentration terms or variable. For a complete system having P phase and C components the total number of variable required = P(C − 1). Total number of variables is given by V = P(C − 1) + 2 (‘2’ stands for temperature and pressure variables) (ii) The number of variables defined by the system itself i.e. number of relations of equilibrium: When a heterogenous system is in equilibrium at constant temperature and pressure, the chemical potential of any component will have the same value in all the P phases. Thus if a system consist of three phases, Say a , b and g, then for any component i, (μi) a = (μi) b = (μi) g If one of them is taken as standard value, then two equations are written as (a) (μi) a = (μi) b = (μi) g (b) (μi) a = (μi)g

Thus for a system of 3 phases, two equations are known for each component. Similarly, for a system of 4 phases, three equations are known for each components. In general, for a system of P phases, (P − 1) equations are known for each components. Hence, for a system having P phases and C components.

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8.8 Engineering Chemistry Number of equations known will be = C (P − 1) So, number of variables defined by the system itself = C(P − 1) E = C(P − 1) Now, degree of freedom of the system having P Phases and C components will be given by F=V−E = [P(C − 1) + 2] − [C(P − 1)] F=C−P+2 This is the statement of Phase Rule, where only three state variable (Temperature, pressure and concentration) are taken into consideration. If one of these two variables (temperature and pressure) does not affect on equilibria then degree of freedom for such a system will be reduced by one and in this case phase rule is called as reduced phase rule and it is represented as F′ = C − P + 1 Reduced phase rule equation.

8.3.3 Utility of Phase Rule | Application of Phase Rule The following are some of the advantages from the study of phase rule. (i) Phase rule helps in classifying various equilibrium states of the system in terms of the number of components, number of phases or the number of degrees of freedom. (ii) The different systems which are having the same value of degree of freedom according to phase rule, they would behave in a similar fashion. (iii) By using phase rule, we can predict the behavior of any system with the changes in the variables such as temperature, pressure and concentration. (iv) Phase rule is applicable to physical and chemical equilibria irrespective of the nature or amounts of the substances. (v) It finds extensive use in the study of heterogeneous systems. This rule helpful in metallurgy and provided us useful information about the complex formation among different components. (vi) By using phase rule, we can predict whether under a given set of conditions, a number of substances taken together would remain in equilibrium as such or would involve inter-conversion or elimination of some of them substances.

8.3.4 Limitations of Phase Rule (i) Phase rule applies only to a single equilibrium state and does not tell us about the number of other possible equlibria present in the system. (ii) The phase rule equation (F = C − P + 2) is applicable only to system having three variables. When the numbers of variables change the equation shall have to be changed accordingly. (iii) Phase rule deals with macroscopic system only and it does not tell us anything about the molecular structure.

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Phase Rule

8.9

(iv) The phase rule does not give any information regarding the time taken for the system to attain equilibrium. (v) The phase rule also does not give any information regarding the amount of any phase under equilibrium. It only considers the number of phase present in the system at equilibrium.

8.4 Phase Diagrams Phase diagram is the complete description of behavior of phases in equilibrium. When the system is in equilibrium, the number of phases that exist together depends upon the conditions of temperature and pressure concentration kept constant or conditions of temperature and composition, pressure being kept constant. The diagrams so obtained giving the conditions of equilibria between various phases of a substance are called as phase diagrams or equilibrium diagrams. The phase diagram contains a number of lines, areas and point by intersection and by such diagrams we are able to know the conditions under which various phases will be present in the system. Let us consider one component system with different number of phase (i) Single phase: When a pure substance is in single phase and having one component. Then according to phase rule equation F=C−P+2 = 1 − 1 + 2 = 2 (Bivariant) It means such a system can be completely describe by using two variables (ii) Two phase: When we are considering one component in two phase are in equilibrium Then degree of freedom F=C−P+2 = 1 − 2 + 2 = 1 (Monovariant) All such type of systems can be completely described by stating only one variables either temperature or pressure. (iii) Three phase: When we are considering one component in three phase are in equilibrium, then the system has no degree of freedom F=C−P+2 = 1 − 3 + 2 = 0 (Invariant) For all such type of system, all variables must be specified fixed.

8.5 One Component System For a one component system, According to phase Rule,

F=C−P+2 C=1 F=3−P

(i) When value of P = 1, then Degree of freedom, F = 2, it means both P and T can be varied independently without disturbing the equilibrium.

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8.10 Engineering Chemistry (ii) When two phases are in equilibrium, value of P = 2. Degree of freedom (F ) = 1, it means pressure can be changed freely if temperature is set or vice-versa. (iii) When three phases are in equilibrium (P = 3). The degree of freedom (F) is equal to zero it means system can only be non-variant in specific or definite conditions of temperature and pressure. Water System Water system is most typical example of one component system, in which the same chemical compound exists in the three phases in equilibrium as Ice water Vapour (solid)

(liquid)

(gas)

These three phases may occur in the three possible combinations of the two phases in equilibrium as (i) water Vapour vapourisation process (liquid)

(gas)

(ii) Ice Vapour sublimation process (solid)

(gas)

(iii) Ice water Fusion process (solid)

(liquid)

The conditions of temperature and pressure at which the various phases can exist have been determined experimentally and summed up in Figure 8.2. These phase diagram of water system is represented as in Figure 8.2. C

Critical pressure

cu io n

or is

Liquid phase

Va p

u on c Fusi

Solid phase

rv e

rve

1 atm O

4.58 mm A′

ati

m bli

rv cu

Triple point Vapour phase

Critical temperature

Vapour pressure (not to scale)

218 atm

B T1 −273°

BP T2

0.0023° 0.0098° 100°

374°

Temperature °C (not to scale)

Figure 8.2 Phase diagram of water system

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Phase Rule

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The phase diagram consists of: (i) Three stable curves OA, OB and OC and a meta stable curve OA′ (ii) Three areas AOC, BOC and below AOB (iii) Definite point, O (a) Areas (i) Area below AOB In this area vapour phase of water exists, According to phase rule F=C−P+2 =1−1+2=2 System is bivariant in this area. It means to show the conditions of existence of this phase two variables temperature as well as pressure are required. (ii) Area BOC In this area solid phase (Ice) exists. Again degree of freedom F = C − P + 2 = 1 −1 + 2 = 2 The system is bivariant and therefore to specify the conditions of as existence of this phase temperature and pressure are to be specified. (iii) Area AOC This area consist liquid (water) phase Degree of freedom F = C − P + 2 =1−1+2=2 Due to bivariant nature of the system in this area both variable (T & P) are to be specified to explain the conditions of existence of liquid phase of water system. (b) Curves (i) Curve OA: It represents liquid-vapour equilibrium liquid Vapour This curve is called as vapourization curve or vapour pressure curve. This curve explain with increase of temperature, the vapour pressure increases. This curve status from the point O the triple point of water (0.0098 °C at 4.58 mm) and extends upto critical temperature (374 °C) at critical pressure (218 atm), beyond which the two phases merge into each other. Here P = 2 and C = 1 According to phase Rule F = C − P + 2 = 1 − 2 + 2 = 1 (Monovariant) Since the degree of freedom is one, hence the system is univariant or monovariant. It means, for any give vapour pressure on curve there is only one value of temperature or vice-versa. At 100 °C, the vapour pressure of water equals the atmospheric pressure (760 mm). This is therefore, called as boiling point of water. Beyond the point ‘A’ (critical temperature) liquid phase of water does not exist. The slope of the curve OA is positive i.e. the vapour pressure of water increases with temperature. It is also predicted by Clausius-Clapeyron equation as

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8.12 Engineering Chemistry ∆H v dP = = + ve dT T (Vg − Vl )

(∵ Vg > Vl )

Where ∆Hv = Change in molar heat of vaporization Vg = Molar volume of water vapours Vl = Molar volume of water liquid T = boiling point of water (ii) Curve OB: It is known as sublimation curve or vapour pressure curve of Ice. It represents equilibrium between solid and vapour phase solid vapour It gives various values of temperature and pressure at which ice and vapour can exist together. This curve starts from point O, the triple point of water and extends upto point B (absolute zero, −273 °C) Here P = 2, C = 1 F=C−P+2=1 Since, the degree of freedom is one, hence the system is univariant i.e. for each temperature that can be one and only one pressure and vice-versa. Here the slope of the curve OB is positive and this is also predicted by the Clausius-clapeyron equation ∆H s dP = = + ve (∵ Vg > Vs ) dT T (Vg − Vs ) Where ∆H s = Charge in molar heat of sublimation Vg = Molar volume of gas phase Vs = Molar volume solid phase. (iii) Curve OC: It is called as melting point or fusion curve and it represents the equilibrium between ice and liquid water at various pressure. solid liquid Here P = 2, C = 1 F=C−P+2=1 Since, the degree of freedom in one, hence the system is univariant. The slope of the curve OC is negative i.e. meeting point of ice is lowered by increases of pressure. It is also predicted by Clasius-Clapeyron equation ∆H f dP = = −ve (∴Vs > Vl) dT T (Vl − Vs ) Where ∆Hf = change in molar heat of fusion. Since the density of ice is less than that of water, so Vs > Vl, dP Hence should have negative sign. This means that the increase of pressure must lower and dT decrease of pressure must increase the freezing point of water.

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Phase Rule

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(iv) Metastable Curve OA′: It represents the liquid water and water vapour in metastable equilibrium. liquid vapour It is vapour pressure curve of super cooled water, and it is possible to cool liquid water below its freezing point without the separation of ice and is shown by dotted line curve OA′; this state is known as metastable state. It is an unstable state because the curve OA′ lies above the curve OB. It means that at the same temperature, the vapour pressure of the metastable supercooled water is higher than the vapour pressure of the stable solid phase. If there is any disturbance in the process equilibrium may disturb and water may suddenly freeze. Here P = 2, C = 1 F=C−P+2=1 The degree of freedom is one hence the system is univariant. (c) Point The point at which all the three phases of water, that is, ice, water and vapour coexist in equilibrium is called triple point. Here, the curves OA, OB and OC intersect with each other at point ‘O’. This point is shown in the phase diagram at point ‘O’. Ice Liquid Vapour Here P = 3, C = 1 F = C − P + 2 =1−3+2 =0 Since the degree of freedom is zero, hence the system is invariant. The temperature and pressure corresponding to this equilibrium are 0.0098 °C and 4.58 mm respectively. The three phases can co-exist in equilibrium under only one set of conditions; even slight changes in the temperature and pressure will shift the equilibrium and the three phases cannot co-exist. Salient features of water system Curve/Area/ Point

Name of the system

Phase equilibrium

No. of phase (P)

Degree of freedom (F)

Curve OA

Vaporization curve

liquid vapour

01 (univariant)

Curve OB

Sublimation curve

solid vapour

01 (univariant)

Curve OC

Fusion curve

solid liquid

01 (univariant)

Curve OA′

Metastable vaporiza- liquid vapour tion curve – Liquid – Solid – Vapour

01 (univariant)

01 01 01

02 (bivariant) 02 (bivariant) 02 (bivariant)

Zero (invariant)

Area AOC Area BOC Area AOB (Below AOB) Point O –

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8.14 Engineering Chemistry

8.6 Two Component System According to Phase Rule F=C−P+2 C=2 F=4−P For a system with two components and one phase degree of freedom will be three. F=4−1=3 Therefore, three variables would be necessary to describe a system i.e. temperature, pressure and composition, all of these variables are required, therefore a three dimensional model is required for presentation which is more complex and difficult to understand we have a 2D planes and therefore we consider the one variable as constant, only two variables are considered. This reduces the degree of freedom of the equilibrium system and phase rule is written as F1 = C − P + 1 This is known as reduced or condensed phase rule. In general, most of experiments are conducted in open vessel where pressure in constant so, we generally expressed phase diagram of two components by using variable temperature and composition.

8.6.1 Eutectic System Eutectic system is a binary system consisting of two substances that are miscible in all proportions in the liquid (solution) phase without reacting chemically. (The term eutectic means easy ‘to melt’.) For example, bismuth–cadmium, lead–silver, KI–water systems, etc. Eutectic Mixture Eutectic Mixture is a solid solution of two or more substances having the lowest freezing point of all the possible mixture of the components, this is taken advantage of in “alloy of low melting point” which are generally eutectic mixture. Eutectic Point Two or more solid substances capable of forming solid solution with each other have the property of lowering each other’s freezing point, and the minimum freezing point attainable corresponding to the eutectic mixture is termed as the eutectic point. For example, in Pb-Ag system, eutectic point is achieved when the composition is 97.4% Pb and 2.6% of Ag. Systems giving rise to eutectic point are known as eutectic systems. The eutectic composition and temperature of two metals and salt-water system is given in Table 8.1. Characteristics of Eutectic Point (i) Eutectic point represents the lowest or limiting temperature at which a liquid phase can exist in the system or it is the maximum temperature upto which a solid phase can exist. (ii) No other mixture containing the two components will have a melting point lower than the eutectic temperature.

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Phase Rule

8.15

Table 8.1 Some eutectic systems S. no.

Components and their melting points

Eutectic composition

Eutectic temperature

1. 2. 3. 4. 5. 6. 7.

Ag (961 °C); Cu (1083 °C) Ag (961 °C); Pb (327 °C) Zn (419 °C); Cd (323 °C) Zn (419 °C); Al (658.7 °C) Bi (273 °C); Cd (323 °C) KCl (773 °C); H2O (0 °C) KI (682 °C); H2O (0 °C)

(71.8% Ag + 29.2 % Cu) (2.6% Ag + 97.4 % Pb) (67% Zn + 33 % Cd) (95.64% Zn + 4.36 % Al) (60% Bi + 40 % Cd) (20% KCl + 80 % H2O) (52% KI + 48 % H2O)

778 °C 303 °C 270 °C 380.5 °C 140 °C -11 °C -22 °C

(i) Eutectic point has precise values of temperature and composition and it represents an invariant system. (ii) If the liquid is cooled to just below the eutectic point, both the components of the eutectic simultaneously solidify without any change in the composition or temperature of the liquid phase. (iii) An eutectic system can maintain its temperature constant for long periods. (iv) When the liquid is cooled below the eutectic point, the components solidify in the form of small crystal intimately mixed with each other which fill in the spaces between the large crystals of the pure components which are already separated out. (v) Eutectic mixtures appreciably contribute towards the strength of the solid structure in case of alloys. (vi) Eutectic are mixtures of the components but not their compounds. Applications of Eutectic System It has wide applications in industries, pharmaceutical science; medical science etc. Some important applications are the following: (i) (ii) (iii) (iv)

Freeze–drying Safety plugs or safety fuses Solders and Freezing mixture.

(i) Freeze–drying The freeze–drying is the complete removal of water from the material, such as food. The following are two advantages of freeze–drying of food materials: (i) The microorganisms require water to survive. If water is removed from food, then it won’t get spoiled for a long period of time due to non-survival of microorganisms. Enzymes also require water to react with food, so dehydrating of food will also stop ripening. Freeze-dried food materials such as fruits, vegetables etc. can be stored for years and can be completely revitalized with a little warm water. After the years, the taste and texture will be very much the same. (ii) Freeze-drying significantly reduce the total weight of the food .Freeze dried food can be easily transported to the military during war. It is also used by astronauts in a space craft. Difference between Dehydration and Freeze Drying Freeze drying is different from ordinary dehydration. In dehydration, water present in the material is removed by evaporation of water from liquid to vapour state by supplying heat energy. The food

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8.16 Engineering Chemistry materials will be decomposed by the heat supplied for evaporation. Moreover the food will lose its taste, texture and vitality. But in freeze drying, the liquid water present in the food material is freezed to solid ice, which is then sublimed under such conditions that the decomposition and volatilization of other constituents is avoided. Principle of the Process Sublimation is the fundamental principle in freeze–drying, that is, the conversion of a solid directly into a gas. Sublimation occurs when a molecule gains enough energy to break free from the molecules around it, just like evaporation. Water will sublime from solid form (ice) to gaseous form (vapour) when the molecules have enough energy to break free but the condition that the solid melting to its liquid form, which normally happens, does not take place. (ii) Safety Plugs or Safety Fuse A safety fuse is a protective device made from a low melting alloy that melts under heat produced either by excess heating or by an excess current in the circuit. They are used to ensure the safe working and avoid accidents. Most commonly used low melting alloys are wood metal, rose metal and fuse wires. (i) Wood metal contain 50% Bi, 25% Pb, 12.5% Sn and 12.5% Cd. Its melting point is about 70 ° C. It is used for making safety plugs for cookers, fire alarms and for boilers and electric fuses etc. (ii) Rose metal contains 50% Bi, 28% Pb and 22% Sn. Its melting point is 88 ° C. It is used for making fuse wires, fire arms and automatic sprinklers. (iii) Fuse wires for small current are made of Pb–Sn alloy and for high current are made of Pb, Sn, Zn, Sb, Cu, Al etc. (a) Safety fuses as plugs are installed in building to protect them against any fire hazards. When a building catches fire, the heat melts fusible alloy plug and water rushes out from the pipe. This controls the fire automatically. (b) They are fitted in the form of plug in steam boilers and pressure cookers. Whenever steam pressure exceeds the limiting value of pressure, safety plug gets over heated and melts, thereby permitting excessive steam to escape out of the boiler. Thus, accident due to overheating can be avoided. (c) Safety fuse in the form of fuse wire is also used for protection the cable in an electrical circuit against damage from any excessive current than normal. This is because when current is exceeded than normal value, fuse wire gets heated upto the melting point of fuse wire thereby fuse wire gets melted and circuit is broken. As a consequence, wire in the circuit gets protected from over-heating. (iii) Solders A solder is an alloy having lower melting point than that of the individual metals that are joined together by melting. It works based on the principle of eutectic mixture freezes sharply at its freezing point; hence, solders have somewhat different compositions from the eutectic so that the freezing occurs over a range of temperatures.

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Phase Rule

8.17

Table 8.2 Solders and their compositions and uses Solder

Composition

Uses

Soft or Tinner’s solder

50% Pb and 50% Sn or 60% Pb It is commonly used for sealing the tin cans. and 40% Sn Half-Half solder 50% Pb and 50% Sn, Pb and Sn It is very common and famous solder in the are equally shared; hence, it is market. It provides a bright surface in finish called half-half solder after soldering. Brazing alloy 92% Sn, 5.5% Pb and 2.5% Cu These are used for soldering of steel joints. Plumber alloy 67% Pb and 33% Sn It is commonly used for soldering the pipes. The alloy containing about 60% Pb is used as solders in electrical wiring.

The capacity of solders depends upon the formation of a surface alloy between the solder and the parts of metals being soldered. Based on desired melting point and the metals to be joined, the solder alloy can be selected. Solders usually contain Pb and Sn as the main components. Important solders and their compositions are shown in Table 8.2. A good solder has the following characteristics (i) Its melting point should be lower than the soldering metals (ii) It should spread easily in liquid form and form homogeneous mixture with the soldering metals. (iii) It should possess good quality of wetting the soldering metals. (iv) Freezing Mixture A mixture of ice and salt is known as freezing mixture; it has been observed that the addition of salt to ice results in considerable lowering of temperature. A good freezing mixture should satisfy the following conditions: (i) Salt must be cheap, highly soluble and have a low cryohydric temperature. (ii) The heat of solution of the salt should be high, that is, solubility of the salt should increase rapidly on increasing the temperature. (iii) Components used should form an intimate mixture on cooling. Normally a mixture of ice and common salt is used as a freezing mixture because common salt is very cheap and easily available. However, it is not a good component for freezing mixture due to the heat of solution of the salt is very low and the heat absorbed is almost due to the heat of fusion of ice. Calcium chloride hexahydrate and ice give an excellent freezing mixture because of very low cryohydric point and high heat of solution. Some important freezing mixtures, their eutectic temperatures and percentage composition are listed in Table 8.3. Table 8.3 Freezing mixtures System NH4Cl and ice NH4NO3 and ice NaNO3 and ice NaCl · 2H2O and ice KI and Ice CaCl2 · 6H2O and ice

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Composition (% of salt in the mixture)

Eutectic temperature (°C)

20.1 43.0 33.3 23.0 52.0 15.2

-16.0 -18.0 -18.1 -22.0 -23.0 -55.9

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8.18 Engineering Chemistry

8.6.2 Lead (Pb) – Silver (Ag) System Both lead and silver are completely miscible in liquid state and form homogeneous mixture without formation of any compound by the mixing of lead and silver. Melting point of pure lead and pure silver are 327 °C and 961 °C respectively. Melting point of both components i.e. solvent is lowered by the addition of solute. On addition of silver in lead, melting point of lead is lowered and vice-versa. The phase diagram of Pb·Ag system consists of the following points (a) Curves AO, BO (b) Areas above AOB, below AO and below OB (c) Eutectic point (O) The phase diagram of lead-silver system is shown in Figure 8.3. X′ B(961°C)

Liquid

Temperature (Not to scale)

Y′

C 303°C

Solid Ag + Liquid

A 327 °C Solid Pb + Liquid

Y O

Solid Pb + Eutectic F Pb = 100% Ag = 0%

D Solid Ag + Eutectic

E Pb = 97.4% Ag = 2.6%

G Pb = 0% Ag = 100%

Composition (Not to scale)

Figure 8.3 Phase diagram of lead-silver system Curves (i) Curve AO: This curve represents the melting point or freezing point curve of Pb by addition of small amount of Ag. Along this curve solid Pb+ liquid melt are in equilibrium. Along this curve, the silver which is added goes into solution while the separation of solid Pb takes place. solid Pb liquid (solution of Ag in liquid Pb)

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Phase Rule

8.19

C = 2, P = 2 F=C−P+1=2−2+1=1 The system is univariant along this curve. So, to specify the conditions of equilibrium between two phases only one variable either temperature or composition is to be specified. (ii) Curve BO: This curve represents the melting point or freezing point curve of Ag by addition of small amount of Pb. Along this curve the lead which is added goes into solution while separation of solid Ag takes place. Along this curve two phases solid silver and liquid melt are in equilibrium. Solid Ag liquid (Solution of Pb in liquid Ag) C=2P=2 F=C−P+1=2−2+1=1 The system is univariant along this curve as two phases solid Ag and liquid are in equilibrium. Only one variable either temperature or composition is to be specified for specify the conditions of equilibrium between two phases. Areas (i) Area above AOB In this area only one phase i.e. liquid melt (solution of Ag and Pb) co-exists F=C−P+1 =2−1+1=2 The system is bivariant in this region. So, to show the existence of this phase in this region two variables temperature as well as composition are required. (ii) Area below AO or Area enclosed between AOC In this area AOC, solid Pb and liquid melt co-exist According to phase rule equation F=C−P+1 = 2 − 2 + 1 = 1 (Univariant) The system is univariant in this region. Only one variable either temperature or composition is required to specify the equilibrium state. (iii) Area below BO or Area enclosed between BOD In this area, solid Ag and liquid melt co-exist. So, according to condensed phase rule equation

The system is univariant. (iv) Area below the line COD

F=C−P+1 =2−2+1=1

(a) Area enclosed between COEF: Solid Pb and eutectic phases are co-exist in this region. F=C−P+1=2−2+1=1 The system is univariant in this region.

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8.20 Engineering Chemistry (b) Area enclosed between ODGE: Solid Ag and Eutectic phases are co-exist in this region. F=C−P+1=2−2+1=1 The system is univariant in this region. Points (i) Point A and Point B: Point A represents the melting point of pure Pb (327 °C) and point B represents the melting point of pure Ag (961 °C) At point ‘A’ Solid Pb liquid Pb P = 2, C = 1 F = C − P + 1 = 1 − 2 + 1 = 0 (Invariant) At point ‘B’ Solid Ag liquid Ag P = 2, C = 1 F = C − P + 1 = 1 − 2 + 1 = 0 (Invariant) (ii) Eutectic Point ‘O’ It is a point where two curve AO and BO meet. At this point, three phases are in equilibrium At point ‘O’ Solid Ag Solid Pb liquid meet (solution of Pb and Ag) F=C−P+1 =2−3+1=0 Hence, system is invariant. This point ‘O’ represents the lowest possible temperature (303 °C) below which a liquid phase cannot exist and beyond which the liquid phase cannot be enriched in either component by freezing out the other component. Such type of liquid mixture of Pb and Ag which has the lowest freezing point corresponding to all other liquid mixtures is called eutectic mixture and corresponding temperature is known as eutectic temperature. At Eutectic point Pb = 97.4%, Ag = 2.6% and Eutectic temperature = 303 °C. Cooling of Melt in the Area Above AOB/Pattinson’s Process for Desilverization of Lead When a liquid melt of certain composition at a point X (or X′) in the area above AOB is allowed to cool, it follows Newton’s Law of cooling, as follows the path XY(or X′Y′) in the phase diagram. Lead crystals start separating when point Y is reached and further follows the path YO as we continue the cooling of liquid melt and correspond silver crystals separate out along the path Y′O and point ‘O’ is reached, this point is known as eutectic point (low melting) and the solid mixture is called eutectic mixture which has a characteristic composition for each system.

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Phase Rule

8.21

8.7 HEAT TREATMENT OF STEEL Heat treatment is the combination of heating and cooling of a metal or alloy in one or more temperature cycles to get desirable physical properties to the metals or alloys. Heat treatment of steel may be carried out under near equilibrium conditions to enhance the ductility or under non-equilibrium conditions to enhance the hardness. During heat treatment, the shape and size of the grains or the compositions of the phase undergoes changes with respect to the microconstituents.. Pure iron is not useful for fabrication of structure component because of its weak mechanical properties. So a non-metallic element i.e. carbon forms alloys with iron to give various type of steel and improves the mechanical properties of the base metal. Due to small size of carbon as compared with iron atom carbon interstitial positions in the lattice formed depends on the crystal structure of iron which in turn depends on the temperature, as pure iron exist in three allotropic modifications of a, g and d forms. The low temperature allotropic form called the α-iron has a body-centered cubic (BCC) structure which is stable upto 910 °C. In the temperature range between 910 °C 1400 °C, g -iron with a face centered cubic (FCC) structure is stable and the high temperature allotropic form d -iron has a BCC structure and it is stable beyond 1400 °C and upto 1535 °C. Plain carbon steel on heating to temperature more than 723 °C and maintained at this temperature for a long time allows the formation of the austenite phase and the dissolution of more carbon in the FCC structure. On slow cooling of the austenite phase transformation of FCC to BCC occurs and the excess carbon forms cementite. Hardening If the steel is quenched by plunging into water or oil to 204°C or a lower temperature, the carbon atom do not have sufficient time to form cementite but remain trapped in the BCC structure. The excess carbon precipitates out in the hot metal and prevents the slipping of the planes. Hence, quenched steel is quite hard and strong but has lower ductility; this heat treatment is called as transformation hardening. This involves the transformation of austenite to martensite or the bainite-phase making the steel hard. Tempering Due to its brittleness, the quenched steel is not useful for construction purpose; hence, quenching is always followed by another heat treatment process called tempering. The quenched steel is tempered by reheating to below the a -iron to g -iron transition temperature. The residual stress and strain are relieved, and the excess of carbon is rejected in the form of e -carbide (Fe2.4C). By tempering the steel becomes tougher and ductile. Tempering is carried out at about 200 °C to make hard steel resistant to abrasion or at higher temperature (~540 °C) to make tough steel capable of withstanding shock loads. Annealing This involves heating and holding the steel at a suitable temperature for some time to facilitate the dissolution of carbon in g -iron in a furnace; steel is softened and becomes ductile and also machinable. However, annealing decreases the hardness and strength of the steel. Annealed hypereutectoid steel contains cementite. It is not soft but can be machined easily. In contrast, annealed hypo-eutectoid steel contains ferrite and is relatively soft and malleable.

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8.22 Engineering Chemistry 1000

900

Temperature (°C)

g + Fe3C a +g

800 a

Spheroidizing

700 Process anneal 600

500

a + Fe3C

0.2

0.4

0.6

0.8

1.0 1.2 Carbon (%)

1.4

1.6

Phase diagram of iron carbon system

8.8 REVIEW QUESTIONS 8.8.1 Fill in the Blanks 1. A homogenous and physically distinct and mechanically separable part of system is called as ____________. [Ans.: Phase] 2. Number of phases when CaCO3(s) heated is ____________. [Ans.: 3 (Three)] 3. The smallest number of independent variable constituents by which the composition of each phase can be expressed in the form of chemical equation is ____________. [Ans.: Components] 4. The number of component when NH4Cl heated in a closed vessel is ____________. [Ans.: One] 5. The point at which the gaseous, liquid and solid phases of the system co-exist in equilibrium is called as ____________. [Ans.: Triple point] 6. Degree of freedom is zero at triple point is called as ____________. [Ans.: Invariant system]

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Phase Rule

8.23

7. At Invariant system, value of F is ____________. [Ans.: Zero] 8. In the expression C = S - R, R is number of ____________. [Ans.: Independent chemical reactions] 9. In case of ionic reaction, the formula for calculating number of component is ____________. [Ans.: C = S – (R + 1)] 10. The point which refers to the temperature and pressure where a liquid and its vapour become identical is called as ____________. [Ans.: Critical Point] 11. The degree of freedom at the triple point of one component system is ____________. [Ans.: Zero] 12. The existence of a solid substance in more than one crystalline form is called as ____________. [Ans.: Polymorphism] 13. In a phase diagram, the crossing of a two-phase equilibrium curve is called as ____________. [Ans.: Transition] 14. The melting point of ice can be lowered by an increase of ____________. [Ans.: Pressure] 15. Eutectic temperature of Pb-Ag system is ____________. [Ans.: 303 °C] 16. Phase diagram of Pb-Ag system is plotted between temperature and ____________. [Ans.: Composition] 17. Mathematical statement of reduced phase rule is ____________. [Ans.: [F′ = C - P + 1]] 18. At Eutectic point the composition of Pb = ____________ and Ag = ____________. [Ans.: = 97.4%, 2.6%] 19. Metals essential for solders are ____________ and ____________. [Ans.: Lead and Tin] 20. In order to get a two metal low melting alloy, ____________ composition is preferred. [Ans.: Eutectic] 21. The fundamental principle of freeze-drying is ____________. [Ans.: Sublimation] 22. Iron –Carbon alloys containing 1.7% of carbon are classified as ____________. [Ans.: Steel]

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8.24 Engineering Chemistry 23. The percentage of carbon in cast iron is ____________. [Ans.: 3%] 24. The most important low melting alloys are ____________ and ____________. [Ans.: Wood metal, Rose metal] 25. A ____________ is a low melting alloy which is used to join two metal pieces together. [Ans.: Solder] 26. Soft solders mainly contain ____________ and ____________. [Ans.: Sn, Pb] 27. An ____________ equilibrium diagram indicates the phase changes during heating and cooling and the nature and amount of structural component of steel and cast-iron exist and any temperature. [Ans.: Iron-carbon] 28. The solubility or carbon in iron depends on the ____________ of iron. [Ans.: Crystal structure] 29. The curve above which the system consist of liquid phase only is ____________. [Ans.: Freezing curve] 30. The curve below which the system consist of solid phase only is ____________. [Ans.: Melting curve]

8.8.2 Multiple-choice Questions 1. The Gibb’s phase rule is given as (a) F = C - P + 2 (c) F = C + P - 2 [Ans.: a]

(b) F = C - P + 1 (d) F = C + P + 2

2. The condensed phase rule is given as (a) F = C - P + 2 (c) F = C + P + 1 [Ans.: d]

(b) F = C - P + 3 (d) F = C - P + 1

3. Which of the following has higher vapour pressure (a) super-cooled water at -5 °C (b) Ice at -5 °C (c) Vapour at -5 °C (d) None of these [Ans.: a] 4. Dissociation of NH4Cl in a closed vessel is a (a) one component, one phase system (b) one component, two phase system

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Phase Rule

8.25

(c) Two component, one phase system (d) Two component, two phase system [Ans.: b] 5. A system consisting of rhombic sulphur, monoclinic sulphur and vapour in equilibrium is (a) Non variant (b) univariant (c) bivariant (d) trivariant [Ans.: a] 6. In the lead-silver system, the percentage of silver present at the eutectic point is (a) 0% (c) 97.4% [Ans.: b]

(b) 2.6% (d) 100%

7. In water system, the three phases exist in equilibrium at (a) 0 °C, 1 atm (b) 0 °C, 4.58mm (c) 0.0098 °C, 1 atm (d) 0.0098 °C, 4.58mm [Ans.: d] 8. Water system is non variant at (a) Melting point (c) Triple point [Ans.: c]

(b) Boiling point (d) Critical point

9. Number of phases present in a mixture of N2 and H2 are (a) 1 (c) 3 [Ans.: a]

(b) 2 (d) 4

10. Number of component present in the following reaction CaCO3(s) → CaO(s) + CO2(g) (a) 1 (b) 2 (c) 3 (d) 4 [Ans.: b] 11. At the eutectic point, a system has (a) The lowest melting point (c) Only two phases [Ans.: a]

(b) The highest melting point (d) Uncertain composition

12. What is degree of freedom at eutectic point (a) 1 (b) 0 (c) 2 (d) 3 [Ans.: b]

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8.26 Engineering Chemistry 13. At eutectic point, composition of Pb-Ag system (a) 100% Pb + 0% Ag (b) 0% Pb + 100% Ag (c) 50%Pb + 50% Ag (d) 97.4%Pb + 2.6% Ag [Ans.: d] 14. For C = 1 and P = 3, the degree of freedom is (a) 0

(b) 1

(d) 3

[Ans.: a] 15. For water system, the value of component is (a) 0 (b) 1 (c) 2 (d) 3 [Ans.: b] 16. Two miscible liquids such as water and alcohol has how many phases (a) 1 (b) 2 (c) 3 (d) 4 [Ans.: a] 17. Two inmiscible liquids such as water and mercury has how many phases (a) 1 (b) 2 (c) 3 (d) 4 [Ans.: b] 18. A system of NaCl-KCl-H2O has how many components (a) 1 (c) 3 [Ans.: c]

(b) 2 (d) 4

19. A binary system consisting of two substances, which are miscible in all proportions in the liquid phase, but do not react chemically is known as (a) Invariant system

(b) Univariant system

(d) Eutectic system

20. The hardest structure that appears on the iron-carbon equilibrium is (a) Ledeburite (b) Pearlite (c) Cementite (d) Ferrite [Ans.: c]

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Phase Rule

8.27

21. In Pb-Ag system, the euetectic temperature is (a) 298 °C (c) 303 °C [Ans.: c]

(b) 300 °C (d) 310 °C

22. The melting point of pure Pb and pure Ag is (a) Pb = 100 °C, Ag = 100 °C (b) Pb = 373 °C, Ag = 298 °C (c) Pb = 327 °C, Ag = 961 °C (d) Pb = 961 °C, Ag = 327 °C [Ans.: c] 23. A system consists of water  vapour, the degree of freedom is (a) 0 (b) 1 (c) 2 (d) 3 [Ans.: b] 24. With increase of pressure, melting point of ice is (a) Increase (b) decreases (c) remains unchanged (d) doesn’t show any behaviour [Ans.: b] 25. For one component system, the maximum number of degree of freedom is (a) 0 (b) 1 (c) 2 (d) 3 [Ans.: c] 26. A binary alloy system having value of component is (a) C = 1 (b) C = 2 (c) C = 3 (d) C = 4 [Ans.: b] 27. Soft or Tinner’s solder consists of mainly (a) Pb and Sn (c) Sn and Ag [Ans.: a]

(b) Pb and Ag (d) Pb and Bi

28. Wood metal alloy containing metal is (a) 50% Bi + 20% Pb + 20% Sn + 10% Cd (b) 20% Bi + 50% Pb + 20% Sn + 10% Cd

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8.28 Engineering Chemistry (c) 50% Bi + 25% Pb + 12.5% Sn + 12.5% Cd (d) 25% Bi + 50% Pb + 12.5% Sn + 12.5% Cd [Ans.: c] 29. Plumber solder used for soldering consist of (a) 67% Pb + 33% Sn (b) 50% Pb + 50% Sn (c) 33% Pb + 67% Sn (d) 25% Pb + 75% Sn [Ans.: a] 30. The principle of freeze-drying is (a) Evaporation (c) sublimation [Ans.: c]

(b) condensation (d) Fusion

8.8.3 Short Answer Questions 1. Define or state phase rule. Ans.: Phase rule can predict qualitatively the effect of temperature, pressure and concentration on a heterogenous system which is in equilibrium and it is assumed that equilibrium is not influence by gravity, electrical or magnetic forces, or by surface action. The degree of freedom is related to number of components (C) and of phases (P) by the phase rule equation. F=C-P+2 2. What is condensed phase rule? Ans.: When the pressure of the system remains constant, then the phase rule equation becomes. F=C-P+1 3. Point out the effect of increase of pressure on the melting point of ice. Ans.: Melting point of ice decreases with rise of pressure as shown in the water system diagram that curve is inclined towards pressure axis. 4. Define phase Ans.: A phase may be defined as any part of a system which is homogeneous in itself, physically distinct, and mechanically separable from other parts of the system. 5. Calculate the, degree of freedom of the following systems (i) Unsaturated solution of NaCl in equilibrium with its vapour. (ii) Na2SO4 in water a closed contained at 32.4 °C Ans.: (i) The system has two phases, solution and vapour and it is a two component system. Then degree of freedom F=C-P+2 =2-2+2=2 So, the system is bivariant,

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Phase Rule

8.29

(ii) The system has two phases, solution and vapour and it is a two component system. Since Temperature is fixed at 32.4 °C. So we will apply condensed phase rule equation to evaluate degree of freedom F=C-P+1 =2-2+1=1 So, the system is univariant. 6. What is the number of phases when CaCO3(s) is heated, Ans.:

CaCO3(s) → CaO(s) + CO2(g) Three phases i.e., two solids and one gaseous. 7. How many components are present when NH4Cl is heated in closed vessel. Ans.: One. 8. Give the number of phases, components and degree of freedom of following (a) Mixture of N2 and H2 contained in a vessel. (b) Ice, water and vapour in equilibrium. (c) An unsaturated sugar solution. (d) A system consisting of NaCl, KCl, and H2O. (e) Dissociation of NH4Cl in a closed vessel. (f) Dissociation of NH4Cl in a closed vessel containing NH3 also. Ans.: (a) P = 1 (gaseous phase) C = 2 (N2, H2) F=C-P+2=2-1+2=3 (b) P = 3 (ice, water, vapour) C = 1 (water) F=C-P+2=1-3+2=0 (c) P = 2 (solid, solution) C = 2 (sugar, water) F=C-P+2=2-2+2=2 (d) P = 1 (solution) C = S - R = 5 - (2 + 1) = 2 F=2-1+2=3 (e) P = 2 (solid, gas)

C = 1 (NH4 Cl)

F=C-P+2=1-2+2=1

(f) P = 2 (solid, gaseous)

C = 2 (NH4 Cl, NH3)

F=C-P+2=2-2+2=2

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8.30 Engineering Chemistry 9. Classify the following into true and metastable equilibria (a) Ice and liquid water at atmospheric pressure and 0 °C (b) water at -3 °C (c) water and its vapours at 20 °C (d) SR and monoclinic SM below 95.5 °C Ans.: (a) True equilibrium (b) Metastable equilibrium (c) True equilibrium (d) True equilibrium (e) Metastable equilibrium 10. What is the criterion of multiphase equilibrium? Ans.: Chemical potential of a component must be same in all phases in equilibrium. 11. What is an invariant system. Ans.: A system in which degree of freedom is zero i.e., no condition is required to be specified to define the system. 12. Give an example of invariant system. Ans.: A system consisting of ice, water and water vapour in equilibrium. 13. Give a mathematical statement of phase rule in the following cases:(a) When temperature, pressure and concentration are variables. (b) When temperature and concentration are variables (c) When concentration and pressure are variables. (d) When concentration, pressure, temperature and magnetic force are variables. Ans.: (a) F = C - P + 2 (b) F = C - P + 1 (c) F = C - P + 1 (d) F = C - P + 3 14. Give the degree of freedom of one component system in the following cases (a) When a single phase exists. (b) When two phases are in equilibrium. (c) When three phases are in equilibrium. Ans.: (a) Two (b) One (c) Zero

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Phase Rule

8.31

15. What is the minimum number of phases that can co-exist in a one component system. Ans.: Three 16. What is triple point. Ans.: A point at which the gaseous, liquid and solid phases of the system co-exist in equilibrium 17. What is the triple point of water system. Ans.: When system is at 0.0098 °C and 4.58 mm Hg pressure. 18. What is the difference between a phase and a state of matter. Ans.: There are three states of matter-solid, liquid and gas. A phase is a sample of matter with definite composition and uniform properties throughout the sample. 19. What is meant by polymorphism. Ans.: The existence of a solid substance in more than one crystalline form. 20. What is meant by transition? Ans.: The crossing of a two-phase equilibrium curve in a phase diagram. 21. What is difference between critical point and triple point, Ans.: Critical point refers to the temperature and pressure where a liquid and its vapour become identical while triple point is the condition of temperature and pressure under which three phases of a substance co exit in equilibrium. 22. What is metastable state. Ans.: The state of super cooled or super saturated solution in which the phase, which is normally stable under the given conditions, does not form, under normal conditions. 23. What is meant by the term eutectic? Ans.: A solid solution of two or more substances having the lowest freezing point of all the possible mixture of the components. 24. What is eutectic composition of Pb-Ag system Ans.: 97.4%Pb and 2.6% Ag. 25. What is the degree of freedom at eutectic point? Ans.: zero 26. State the condition in which two substances can form a simple eutectic. Ans.: The two substance must (i) Be completely miscible in the liquid state, but immiscible in the solid state (ii) Not chemically react with each other 27. Justify the statement the eutectic is a mixture and not a compound. Ans.: Eutectic is a mixture of two solids, which exists at the lowest melting point. Since eutectic is completely immiscible in the solid state, so it is a mixture and not a compound.

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8.32 Engineering Chemistry 28. What is the eutectic temperature of Pb-Ag system. Ans.:Euetctic temperature: 303 °C 29. Name two systems which form eutectic mixtures, Ans.: (i) NH4Cl +ice (ii) KNO3 + ice 30. Which metals form the essential constituents of solders and what is the composition of tinman’s solder? Ans.: Lead and Tin are the essential constituents of solders, Composition of tinman’s solder is 66% Sn + 34% Pb 31. What is a phase diagram Ans.: Phase diagram is obtained by plotting concentration versus temperature. 32. Why do study phase diagram. Ans.: To predicts whether a triple point or eutectic alloy or solid solution is formed on cooling a homogenous liquid mixture containing two metals.

8.8.4 Solved Numerical Problems (1) Calculate the number of phases, components and degree of freedom in the following systems (i) CaCO3(s) → CaO(s) + CO2(g) (ii) H2O(s) → H2O(g) + H2O(l) (iii) An aqueous solution of NaCl and Na2SO4 Solution (i) CaCO3 (s) → CaO(s) + CO2(g) Number of phases = 3 [i.e. CaCO3(s), CaO(s), CO2(g)] Number of component, C = S - R S = 3 (number of species) R = 1 (Number of relation) C = S - R = 3 - 1 = 2 Component system So, according to phase rule F = C - P + 2 = 2 - 3 + 2 = 1 (univariant) (ii) H2O(s) → H2O(g) + H2O(l) Number of component = 1 (H2O only) Number of phases = 3 (solid, liquid, vapour) So, degree of freedom F = C - P + 2 = 1 - 3 + 2 = 0 (Invariant)

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(iii) An aqueous solution of NaCl and Na2SO4 Number of phases, P = 1 (i e solution) Number of component, C = 3 (i.e NaCl, Na2SO4, H2O) So, degree of freedom F=C-P+2=3-1+2=4 (2) Write down the number of component, number of phases and calculate the degree of freedom for the following equilibria:(i) N2(g) + 3H2(g) → 2NH3(g) (ii) Fe(S) + H2O(g) → FeO(s) + H2(g) (iii) H2O(g) → H2(g) + ½O2(g) Solution (i) Chemical species present are N2(g), H2(g) and NH3(g) Number of component C = S - R S = 3 [Number of species N2, H2 and NH3] R = 1 [Number of relation] C = S - R = 3 - 1=2 Number of phases P = 1 (gaseous phase) Degree of freedom, F = C - P + 2 =2-1+2=3 (ii) Chemical species present are Fe(s), FeO(s), H2O(g), H2(g) Number of components, C = S - R =4-1=3 Number of phases = 3 (Two solids and one gaseous phase) Degree of freedom, F = C - P + 2 =3-3+2=2 (iii) Number of species present are H2O(g), H2 (g), and O2 (g)

Number of components C = S - (R + 1)

= 3 - (1 + 1) =1

Number of phases, P = 1 (gaseous phase)

Degree of freedom, F = C - P - 2

=1-1+2=2 (3) Explain KCl-NaCl-H2O is a three component system while NaBr-KCl-H2O is a four component system.

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8.34 Engineering Chemistry Solution

First system KCl-NaCl-H2O

Total Number of species, S = KCl, NaCl, H2O, K+, Cl−, Na+ (dissociation of water is neglected)

S = 6

Number of independent relations, R = 2

KCl  K+ + Cl− Nacl  Na+ + Cl− Thus, C = S - (R + 1) = 6 - (2 + 1) = 3

In the second system,

Total number of species, S = 9

KCl, NaBr, H2O, NaCl, KBr, Na+, K+, Cl−, Br−

The number of independent reactions, R = 4

KCl  K+ + Cl− NaBr  Na+ + Br− Na+ + Cl−  NaCl K+ + Br−  KBr C = S - (R + 1) = 9 - (4 + 1) = 4 It is a four component system.

8.8.5 Descriptive Questions Q.1 State phase rule and explain the significance of the term involved. Illustrate with suitable examples. Q.2 Define the terms with suitable example.

(i) Phase (ii) Component (iii) Degree of freedom (iv) Triple point (v) Eutectic mixture (vi) Condensed phase rule

Q.3 Draw a phase diagram for one component water system. Label it and discuss the importance of various points, lines and areas at equilibrium. Q.4 What is meant by triple point of water? Why is it different from the normal melting point of ice? Q.5 Differentiate between True equilibrium and Metastable equilibrium?

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Phase Rule

8.35

Q.6 Derive Gibb’s phase Rule equation. Q.7 Justify the statement ‘The Eutectic is a mixture and not a compound.” Q.8 What is meant by eutectic point? Explain how can the eutectic point be calculated ? Discuss the Pb – Ag system. Q.9 What is the number of phases in the following systems? (i) Saturated solution of NaCl

(ii) Mixture of rhombic and monoclinic sulphur

(iii) Mixture of O2 and N2

(iv) Mixture of benzene and water

Q.10 Water system is a respresentative system for explaining phase rule and phase equilibria. Explain. Q.11 Discuss a typical one-component system from the stand point of phase rule. Q.12 Explain why KCl-NaCl-H2O should be regarded as a three component system: whereas KClNaBr-H2O should be regarded as a four component system. Q.13 What is condensed phase rule? When is it applied? Q.14 Define and explain the term components of a system with suitable examples. Q.15 Discuss the salient feature of Fe-C system. Q.16 Define the various curves involved in water system with a neat and sketched diagram, why is the fusion curve in the phase diagram of water system inclined towards the pressure axis? Explain Q.17 What is a triple point? Explain triple point with reference to water system. Q.18 Explain the terms (i) Stable equilibrium, (ii) Metastable equilibrium, Q.19 Derive phase rule equation viz. F = C - P + 2, why reduced phase rule is applicable in Pb-Ag system. Q.20 Calculate P, C and F in the following cases (i) NH3(g) at 42 °C (ii) An emulsion of oil in water at 2 atm and 70 °C (iii) SR  SM at the transition temperature (iv) Pure crystal of CuSO4 5H2O (v) Water system at 4.578 mm of Hg and at 0.0098 °C

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PHOTOCHEMISTRY

9.1 INTRODUCTION Photochemistry is the branch of chemistry which deals with the effect of light on chemical systems. A photochemical reaction is caused by the absorption of electromagnetic radiation, especially ultraviolet (UV)—visible radiations. Photochemical reactions proceed differently from thermal reactions; hence they have a lot of importance in organic and inorganic chemistry. The photochemical path offers advantages over thermal methods of forming thermodynamically disfavoured products by overcoming large activation barriers and the reaction will be completed in a short period of time. Hence, many thermal reactions have photochemical counterparts. Some examples are photosynthesis, formation of Vitamin D with sunlight in the body and ozone formation when oxygen is exposed to sun light.

9.2 LIGHT SOURCE IN PHOTOCHEMISTRY Photochemists typically work in only a few sections of the electromagnetic spectrum. The most widely used sections are UV and visible regions. The main source of light is the sun, mercury (high, medium and low pressure) lamps, sodium lamps, halogen lamps, LASER, etc. Different lamps and their intensity ranges are shown in Table 9.1. Table 9.1 Different lamps and their intensity ranges Name of the lamp

Wavelength range (nm)

The sun Low-pressure mercury (5–10 atm) High pressure Hg lamps (100 atm) (Highly expensive and easily damaged) Low- and high-pressure sodium lamps High power light emitting diodes (Very narrow and intensive emission, long life time)

300–1400 185–254 360–600 600 400–650

The common types of reactors, vessels and apparatus used for photochemical reactions are irradiated flasks (used for external irradiation), immersion-well reactors (here, the lamp is surrounded by the reaction solution), falling film apparatus and photo-microreactors (similar to falling films, but easy to handle). In all these cases, the lamp usually requires cooling to avoid its overheating.

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9.2 Engineering Chemistry Solid or liquid optical filters may be used to restrict the irradiation wavelength, the glass acts as a good solid filter. The different glasses used for irradiation at different wavelengths are as follows: (i) Quartz glass to irradiate at 254 nm (ii) Pyrex glass to irradiate at 300 nm (iii) Normal lab glass to irradiate > 350 nm

9.3 LAWS OF PHOTOCHEMISTRY Photochemical reaction is initiated by electronically excited molecules or atoms produced by the absorption of electromagnetic radiation, usually visible or near the UV region. Photochemical reactions are governed by three basic principles. They are as follows: (i) Grotthuss–Draper law (ii) Stark-Einstein law of photochemical equivalence (iii) Beer-Lambert law

9.3.1 Grotthuss–Draper Law or The First Law of Photochemistry This law states that light must be absorbed by a chemical substance for a photochemical reaction to take place. Photoexcitation is the first step in the photochemical process, where the reactant is elevated to a state to higher energy an excited state. The Grotthuss–Draper law is also known as the first law of photochemistry. When light pass through any substance, only a fraction of the incident light which is absorbed by the substance brings about a chemical change and the reflected and transmitted light do not produced any such affect. Hence, it is important to remark that in all light radiations, which are absorbed, the reaction systems do not cause an affect in producing the final product. In some cases, the absorbed light is reemitted as radiations of the same belong to a different frequency. The Grotthuss–Draper law is purely qualitative assumption; it does not explain the relation between the amount of light absorbed by the system and the number of molecules which have reacted and give a final product.

9.3.2 Stark-Einstein Law or Photochemical Equivalence Law According to this law, for each photon of light absorbed by a chemical system, only one molecule is activated for the subsequent reaction. This photoequivalence law was derived by Einstein during his development of the quantum theory of light; hence it is known as the Stark-Einstein law. No more than one molecule is activated for the photochemical reaction, as defined by the quantum yield. According to this law, the part of a light induced on the system the primary process can occur this is the initial chemical change that results directly from the absorption of light. However, in most photochemical reactions, the primary processes are usually followed by a secondary process due to normal interactions between the reactants and they are not required to absorb any light radiation. As a result, such reactions do not obey the one quantum–one molecule reactant relationship. The law is further restricted to conventional photochemical processes using the light source of moderate and high intensity sources; those used in flash photolysis and laser experiments are known as biphotonic processes—the absorption by a molecule of a substance of two photons of light.

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Photochemistry

9.3

9.3.3 Beer-Lambert Law The Beer-Lambert law proposed that the absorption of light or single wave length is related to the exponential to the thickness of the absorbing compound and the optical path of light thus:

I t = I o e − an (1)

where Io = intensity of incident light It = intensity of transmitted light a = constant n = thickness of absorbing material. Equation (1) is subsequently modified that thus the Beer-Lambert law of light absorbing is formulated as follows: I = I o e − ∈cl I = e −∈cl Io I log o = ∈ cl (or ) A = ∈ cl I

9.4 PHOTOPHYSICAL AND CHEMICAL PROCESSES After absorbing the energy, the excited molecule may undergo different physical and chemical processes.

9.4.1 Photophysical Process The excited or energised molecule may return to its initial state by any of the following physical processes: (i) The molecule can release the excitation energy by emitting radiation through fluorescence or phosphorescence. (ii) The absorbed energy may transfer to some other molecule to colloids, without emitting light or giving product. (iii) An electron in the atom or molecule may absorb so much energy that it may escape from the atom or molecule, leaving behind the positive M+ ion by photoionisation. Jablonski Diagram The photophysical process can be easily explained by the Jablonski diagram. Once a molecule that has absorbed energy in the form of electromagnetic radiation goes to excitation state, while coming back to the ground state, a number of paths may follow. Quantum mechanics explains internal conversion of energy as a transfer of excess electronic energy into excess vibrational energy of a lower electronic state, and followed by dissipation of vibrational energy into the surroundings as heat. New jabolonski diagram giving a glance about photo physical process shown in Figure 9.1. Internal Conversion Internal conversion is an intermolecular conversion of molecules which possess to a lower electronic state without emitting radiation. The higher excited singlet states (S1, S2, S3, … generally denoted as Sn)

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9.4 Engineering Chemistry and the lowest energy triplet states (T1, T2, T3, …) perform a crossover of two states with the same multiplicity, that is, singlet to singlet or triplet to triplet state. The internal conversion is more efficient when two electronic energy levels are close enough that two vibrational energy levels can overlap in between S1 and S2. Fluorescence The internal conversion can also occur in between S1 and S0 (lowest energy or ground state) and is much slower, allowing time for the molecule to emit a photon or loss of energy from a higher excited state. This is known as fluorescence. Inter-system Crossing The internal conversion from S1 to S0 is due to the vibrational levels of the ground state overlap with the first excited state for some molecules, which leads to fast deactivation. Inter-system crossing is a radiation-less process involving a transition between two different multiplicities—that is S1 (singlet) to T1 (triplet) electronic states. The probability of inter-system crossing is due to the overlapped vibration levels of the two singlet states. This is commonly observed in molecules containing heavy atoms such as iodine or bromine. The spin and orbital interaction increases, and the spin becomes more favourable; paramagnetic species also enhance inter-system crossing, which consequently decreases fluorescence. Phosphorescence The emission of photon by internal conversion of electron from T1 to S0 is known as phosphorescence. The triplet states (with parallel spins) interact more strongly than singlet states (with opposing spins), the energy difference of T1 − S0 is less than S1 − S2. Hence, phosphorescence occurs at longer wavelengths than fluorescence. Important photo physical processes and their transitions shown in Table 9.2. Table 9.2 Photo physical processes and their transitions Photophysical process

Transition occurs

Light absorption (excitation) Internal conversion Vibrational relaxation Inter-system crossing Fluorescence phosphorescence Non-radiative decay

S0 − Sn or Tn Sn − S1 or Tn − T1 Sn* − Sn S1 − T1 S1 − S0 T1 − S0 S1 − S0 or T1 − S0

9.4.2 Photochemical Process A photochemical reaction is a chemical reaction initiated by the absorption of energy in the form of light. Primary Photochemical Process If the excited molecule (M*) reacts, it may undergo any of the following chemical processes: (i) Photodissociation (ii) Inter-molecular rearrangement (iii) Reaction with another molecule

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Photochemistry A = Photon absorption F = Flourocense P = Phosphorosence IC = Internal conversion ISC = Inter system crossing V.R = Vibration relaxation S = Singlet state T = Triplet state

Sn V.R S2 IC Energy

9.5

ISC

S1 F

T2 IC T1

Electronic energy level diagram

Figure 9.1 Jablonski diagram showing photophysical process Photodissociation Dissociation of the molecules of a substance is caused by absorption of radiation energy into atomic and/or molecular fragments. M → M * → M + + n+ Photoisomerisation The conversion of a molecule into its isomer with the same number and types of atoms but a different structural arrangement is called photoisomerisation. N−a→N−b Inter-molecular Rearrangement By irradiating with light, the molecule will absorb the light energy and will rearrange. A−B−C→C−B−A Secondary Photochemical Process The secondary process may occur upon completion of the primary step. Several examples of such process are described here: (i) Ozone formation: O2 → 2O * 2O * + 2O2 → 2O3 (ii) Destruction of ozone in the upper stratosphere (iii) Chain reaction

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9.6 Engineering Chemistry

9.5 QUANTUM YIELD AND QUANTUM EFFICIENCY The efficiency of the photochemical reaction is expressed by the quantum efficiency or quantum yield. It is a measure of the number of product molecules formed by the quantum of energy absorbed from each photon. f is defined as the number of molecules reacted for every number of quanta of light absorbed: f =

Number of molecule reacted Number of quantum light absorbed

The concept of quantum efficiency was first introduced by Einstein as each quanta of light absorbed least formation of product. Though f = 1, in practice, it can be observed from 10−2 to 107. Low quantum is absorbed in cases where the deviation of the molecules takes place before they form the product. The deviation may take place by collisions of excited molecules split into other excited molecules or non-excited molecules in cases where the primary photochemical process gets reversed. The dissociation of molecule takes place and the dissociated fragments may be recombined to form the original molecule. High quantum efficiency is observed in free radical reactions. Photochemical free radical energy is absorbed only in the chain initiation step to the formation of chain initiation-free radical, but will propagate the reaction in the propagation step without absorbing any energy. This process is continuous until the product is formed in the termination step.

9.6 PHOTOSENSITISATION Photosensitisation is a process wherein an electronically excited molecule transfers its energy to another non-radioactive molecule. After transferring energy, the other molecule gets excited and undergoes photochemical change. This process is called photosensitisation. Photosensitisation may be inter-molecular or intra-molecular. The initially excited molecule D* is designated as donor and the non–radioactive molecule is designated as the acceptor. This process is represented by this scheme: D hv → D * Excited state A + D* → A * + D (photosensitisation) There are two types of mechanisms postulated for non-radioactive energy as follows: (i) Long-range transfer by dipole-dipole interaction (ii) Short-range transfer by change interaction

9.7 PHOTODYNAMIC THERAPY Cancer can be defined as a class of disease characterised by uncontrolled growth of a group of cells beyond their normal limit and invasion to adjacent healthy tissues, which involves dynamic changes in the genome. Photodynamic therapy (PDT) has emerged as a promising non-invasive chemotherapeutic technique for the treatment of cancer which uses light to activate the drug molecule (photosensitiser)

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Photochemistry

9.7

to produce reactive singlet oxygen species in the presence of molecular oxygen to damage the cancer cells. Here, cancer cells are selectively targeted and damaged by red light of 600–800 nm wavelength (photospectral window) by leaving the healthy cells unaffected. In PDT, the drug is first injected into the patient’s body and sufficient time is given to accumulate the drug inside the tumour. Subsequently, light is selectively exposed to the tumour to activate the photosensitiser and to kill the cancer cells. Three essential components for the PDT are photosensitiser, light of a particular wavelength and molecular oxygen. A photosensitiser is a biocompatible light sensitive molecule which can be photoexcited by a particular wavelength of light. In the excited state, it can transfer energy to the molecular oxygen to generate reactive oxygen species like singlet oxygen.

Photosensitizer (excited state)

Tissue oxygen Photosensitizer Free radicals singlet oxygen Light source

Light

Photosensitizer (ground state)

Cell Death Oxygen

Cellular toxicity

The photosensitiser used in PDT is mostly the organic molecule. An ideal photosensitiser suitable for PDT application must fulfil the following criteria. First, the dark toxicity should be as less as possible. Second, it should have an excited absorption band at visible wavelength, preferably in the range of 600–800 nm which is known as the PDT window. Shorter wavelength has less tissue penetration and often leads to skin photosensitivity. Longer wavelength of light lowers the quantum yield of triplet formation which hinders energy transfer to the ground state oxygen molecule to excite it to the singlet state. Third, the photosensitiser should have good aqueous solubility and its excretion from the patient’s body should be rapid to avoid side effects.

9.8 IMPORTANT PHOTOCHEMICAL REACTIONS Some examples of photochemical reactions are as follows: Photosynthesis (i) Plants use solar energy to convert carbon dioxide and water into glucose and oxygen with the help of chlorophyll. (ii) Vitamin D is formed when the human body is exposed to sunlight. Polymerisation Many free radical polymerisation reactions start by photoinitiators, which decompose upon absorbing light to produce the free radicals.

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9.8 Engineering Chemistry Photodegradation In many drug bottles, the first precaution is to preserve the drug in a cool and dark place. This is essential to avoid the drug from photodegradation. For example, poly vinyl chloride medicine bottles. Photodynamic Therapy Singlet oxygen is generated by the photosensitised drug when it is exposed to light; it destroys tumours without affecting normal healthy cells. Organic Photochemistry (i) Many organic reactions are initiated by light to give final products with low cost. For example, the formation of cyclo compounds from alkenes, Zimmerman’s di-pi-methane rearrangement, etc. (ii) Industrial preparation of benzyl chloride by the gas-phase photochemical reaction of toluene with hv + • chlorine. Cl 2 → 2Cl C6H5CH3 + Cl• → C6 H 5 CH 2• + HCl C6 H 5 CH 2• +Cl• →C6 H 5 CH 2 Cl (iii) Mercaptan can be prepared by the photochemical addition of hydrogen sulphide to alpha olefins. Inorganic and Organometallic Photochemistry Many coordination and organometallic compounds are also photoactive and involve cis–trans isomerisation, dissociation of ligands in presence of light. For example, tetra hydro furan solution of molybdenum hexacarbonyl gives the THF complex in presence of UV light, which is synthetically useful. MO(CO) 6 + THF → [MO(CO)5(THF)] + CO Like this reaction, the photolysis of iron pentacarbonyl affords diiron nonacarbonyl. Luminescence Emission of light by a substance, which is not due to heating of the substance, is called luminescence. Therefore, it is a form of cold body radiation. Bioluminescence In fireflies, an enzyme in the abdomen catalyses a reaction that produces light. Chemical Luminescence The light produced as a result of chemical reaction is called chemiluminescence. For example, [A] + [B] → product + light Luminol + H2O2 (B) → 3-aminophthalate + light Due to the reaction of luminol with H2O2 in the presence of catalyst, it involves in excitation of molecule in electronic energy levels and produces 3-aminophthalate and light.

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Photochemistry

9.9

Photovoltaic Cell Photovoltaic (PV) cells (or solar cells produce electricity by photoelectric effect; hence, PV cells are the building blocks of all PV systems because they are the devices that convert sunlight to electricity. PV cells are made of semi-conducting materials with different sizes and shapes. They are connected together to form PV molecules that may be up to several feet long and a few feet wide. Molecules, in turn, can be combined and connected to form PV arrays of different size and power output; such as electrical connections, mounting hardware, power-conditioning equipment and batteries that use solar energy when the sun is not shining. When light passes on a PV cell, it may be reflected or absorbed; but only the absorbed light generates electricity. The energy of the absorbed light is transferred to electrons in the atoms of the PV cell semiconductor material, and these electrons escape from their normal position in the atoms and become a part of the electrical flow in an electrical circuit. Molecular Photochemistry This is the study of artificial assemblies of two or more molecules to understand the biological process and the design of artificial systems capable of performing of useful functions. The main three types of supramolecular systems in the area of co-ordination chemistry are as follows: (i) Second-sphere coordination compounds: For example, hexacyano cobalt (III) anion with poly ammonium macro cyclic receptors. The complex is associated with other species by electrostatic interaction, hydrogen bonds or other intra molecular forces. (ii) Cage-type co-ordination compounds: This refers to complexes in which the metal ion is encapsulated in a single polydentate ligand. For example, cage-type cobalt (III) complex. (iii) Molecular building blocks linked via bridging units by means of covalent or co-ordination bonds.

9.9 REVIEW QUESTIONS 9.9.1 Fill in the Blanks 1. Crossover of electronically excited molecule two states with same multiplicity is called ___________. [Ans.: Internal conversion] 2. ________ can prevent the drugs from photodegradation. [Ans.: Poly vinyl chloride medicine bottles] 3. Vision is initiated by a photochemical reaction of ________. [Ans.: Rhodopsin] 4. When the excited molecule is breaks into its atomic or molecular fragment, it is called ____________. [Ans.: Photodissociation]

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9.10 Engineering Chemistry 5. Who introduced the quantum efficiency concept? [Ans.: Einstein]

9.9.2 Multiple-choice Questions 1. _____________ is concerned with chemical effects of light. (a) Photochemistry (b) Photolysis (c) (a) and (b) (d) None of these [Ans.: a] 2. The law that explains thickness of absorbing light and optical path of light is (a) Bear-Lamberts law (b) Grotthuss–Draper law (c) Stark-Einstein law (d) None of these [Ans.: a] 3. Electronically excited molecules that return to ground state with same multiplicity are called (a) Phosphorescence (b) Fluorescence (c) Photosensitisation (d) Luminescence [Ans.: b] 4. Second-sphere co-ordination compounds belongs to (a) Intra molecular photochemistry (b) Inter molecular photochemistry (c) Supramolecular photochemistry (d) Super molecular photochemistry [Ans.: c] 5. Light formed from chemical reaction is called (a) Luminescence (b) Bioluminescence (c) Chemiluminescence (d) All the above [Ans.: c] 6. Fire flies are an example of (a) Bioluminescence (c) Photosynthesis [Ans.: a]

(b) Photolysis (d) Photoisomerisation

9.9.3 Short Answer Questions 1. What are basic laws of photochemistry? Ans.: The photochemical reactions are governed by the following two basic principles: (a) Grotthuss–Draper law (b) Stark Einstein law of photochemical equivalence.

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Photochemistry

9.11

2. Define photochemistry. Ans.: Photochemistry is a branch of chemistry which is concerned with the effect of chemical reactions caused by absorption photon from UV-visible and IR radiation. 3. Explain inter-system crossing. Ans.: Inter-system crossing is a radiation-less process involving transition between two electronic states with different multiplicities. 4. Describe photosensitisation. Ans.: It is a process wherein an electronically excited molecule transfers its energy to nonradioactive molecule to another type of molecule. After excitation, it undergoes a chemical change called photosensitisation. 5. What is meant by photovoltaic cell? Ans.: A photovoltaic cell or solar cell is a device in which electricity is produced by photoelectric effect. 6. Name the types of supramolecular systems. Ans.: Supramolecular systems are of three types as follows: (a) Second-sphere co-ordination compounds (b) Cage-type co-ordination compounds (c) Molecular building blocks linked via bridging units by means of covalent or coordination bonds.

9.9.4 Descriptive Questions Q.1 Explain the laws of photochemistry. Q.2 Describe quantum yield and quantum efficiency of photochemistry. Q.3 Explain the photochemical process in photochemistry. Q.4 Explain the photosensitisation process in detail. Q.5 Give an account of electronically excited states in photochemistry with an energy level diagram. Q.6 Explain the photochemical reactions in photochemistry. Q.7 Give a detailed account of photovoltaic cell. Q.8 Describe supramolecular photochemistry.

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Surface chemistry

10.1 INTRODUCTION “Surface chemistry is the branch of chemistry which deals with the study of the phenomenon occurring at the surface or interface, that is, at the boundary separating two bulk phases.” We know that number of phenomena occur at the interface, for example, dissolution, crystallization, corrosion, electrode processes, etc.

10.2 ADSORPTION It is observed that when certain solids like charcoal, silica, etc., are added into a closed vessel containing some gas or aqueous solution of a solute, the molecules of the gas or the solute are attracted towards the surface of the solid and then retained on the surface. As a result, the concentration of the gas or the solute on the surface of the solid increases. This is known as adsorption. Hence, it may be defined as follows: (i) Adsorption: The phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or a solid resulting into a higher concentration of the molecules on the surface is called adsorption. (ii) Adsorbate: The substance (gas or vapour or the solute) that is adsorbed on the surface (liquid or solid) is called adsorbate. (iii) Adsorbent: The substance (liquid or solid) that adsorbs either gas or vapour or the solute is called adsorbent. (iv) Desorption: The removal of adsorbed substance from the surface is called desorption. It is generally brought by heating or reducing the pressure. (v) Occlusion: The adsorption of gases on the surface of metals is called occlusion. As greater is the surface area of the adsorbent, greater is the adsorption, therefore, finely divided metals or substances having porous structure, for example charcoal, silica gel, clay, etc act as excellent adsorbents. Examples of adsorption: (i) Adsorption of solute by charcoal: When aqueous solution of raw sugar (which has yellowish brown colour) is shaken with animal charcoal and then filtered, the filtrate is colourless, which gives white crystals of sugar on crystallization. This is because the brown colouring substances are adsorbed by animal charcoal.

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10.2 Engineering Chemistry (ii) Adsorption of moisture by silica gel: When silica gel is placed in a closed vessel containing moist air, the air becomes dry after some time. This is because the water molecules are adsorbed on the surface of silica gel.

10.2.1 Mechanism of Adsorption We know that molecules in the interior of a liquid are completely surrounded by other molecules on all sides, and hence intermolecular forces of attraction are exerted equally in all directions. However, a molecule at the surface of a liquid is surrounded by large number of molecules (Figure 10.1). Because the unbalanced, inward forces of attraction on the surface of liquid or solid have the property to attract and retain the molecules of a gas or solute on their surfaces.

(a)

(b)

Figure 10.1 Molecules at the surface experiencing a net inward force of attraction in case of (a) liquid and (b) solid

10.2.2 Adsorption is Exothermic During adsorption process, the residual forces on the surface of the adsorbent decreases; it means surface energy decreases. This decrease that appears in the form of heat is called heat of adsorption. Hence, adsorption is an exothermic process, that is, ∆Hadsorption is always negative. “The amount of heat evolved when 1 mole of any gas is adsorbed on a solid adsorbent surface is called “heat of adsorption or enthalpy of adsorption.”

10.2.3 Difference between Adsorption and Absorption Adsorption is different from absorption. The former refers to the attraction and retention of the molecules of a substance on the surface only, while the latter involves distribution of substance that is uniformly distributed over the liquid or the solid. When both adsorption and absorption take place simultaneously, the process is called sorption. Representation of the three processes is shown in Figure 10.2, and the differences between adsorption and absorption are given in Table 10.1. Surface concentration

Uniform distribution

Adsorption and absorption

Adsorbent

Adsorbate

(a)

(b)

(c)

Figure 10.2 (a) Adsorption process (b) Absorption process and (c) Sorption process

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Surface Chemistry

10.3

Table 10.1 Differences between adsorption and absorption Adsorption

Absorption

1. It is a process in which molecules of a substance are In this process, molecules of a substance are assimilated on the surface of a solid (or liquid). uniformly distributed throughout the body of a solid (or liquid). 2. It is a surface phenomenon, that is, it occurs only at It is a bulk phenomenon, that is, it occurs throughout the surface of the adsorbent. the body of the material. 3. It is a fast process. It is a slow process. 4. Equilibrium is attained easily. Attainment of equilibrium takes some time. 5. It depends upon surface area of adsorbent. It simply requires the porous structure of the substance.

10.2.4 Examples of Adsorption, Absorption, and Sorption (i) When ammonia gas is placed in contact with charcoal, it gets adsorbed on the charcoal, whereas when ammonia gas is placed in contact with water, it gets absorbed into the water, giving NH4OH solution of uniform concentration. (ii) Dyes gets adsorbed as well as absorbed in the cotton fibers, that is, sorption takes place.

10.2.5 Positive and Negative Adsorptions (i) When the concentration of the adsorbate is more on the surface of the adsorbent than in the bulk, it is called positive adsorption. E.g.: When concentrated solution of KCl is shaken with blood charcoal, it shows positive adsorption. (ii) When the concentration of the adsorbate is less on the surface of the adsorbent than in the bulk, it is called negative adsorption. E.g.: When a dilute solution of KCl is shaken with blood charcoal, it shows negative adsorption.

10.2.6 Classification of Adsorption There are two main types of adsorption recognized: (i) Physical adsorption or physisorption or van der Waal’s adsorption (ii) Chemical adsorption or chemisorption or Langmuir adsorption or activated adsorption (i) Physical adsorption: In such type of adsorption, the gas molecules (adsorbed) are held on the surface of a solid by van der Waal’s forces without resulting into the formation of any chemical bond between adsorbate and adsorbent. It is called physical adsorption. (ii) Chemical adsorption: In such type of adsorption, the gas molecules are held on the surface of a solid by forces similar to those of a chemical bond (ionic or covalent). This type of adsorption results in the formation of chemical bond between adsorbate and adsorbent; it is called chemical adsorption. The differences between physical and chemical adsorptions are given in Table 10.2.

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10.4 Engineering Chemistry Table 10.2 Differences between physical and chemical adsorptions Physical adsorption

Chemical adsorption

1. The forces operating in these cases are weak van der Waal’s forces. 2. The heat of adsorption is low, i.e., about 20–40 KJ mol−1. 3. No compound formation takes place. 4. It occurs at low temperatures. 5. This process is reversible in nature, and adsorbate can be recovered from the adsorbent by lowering the pressure. 6. It does not require any activation energy. 7. It forms multimolecular layer. 8. It increases with increase of pressure. 9. It increases with increase in the surface area of the adsorbent. 10. The equilibrium is established rapidly. 11. It is not very specific in nature.

he forces operating in these cases are similar to those T of a chemical bond. The heat of adsorption is high, i.e., 40–400 KJ mol−1. Surface compounds are formed. It can occur at high temperatures. This process is irreversible in nature because adsorbate and adsorbent are strongly held by chemical force. It requires activation energy. It forms unimolecular layer. It also increases with increase of pressure. It also increases with increase in the surface area of the adsorbent. Establishment of equilibrium requires time. It is highly specific in nature.

10.2.7 Factors Affecting the Adsorption of Gases by Solids The amount of gases adsorbed on the surface of solid depends upon the following factors: (i) Nature and surface area of the adsorbent: It is observed that the extent of adsorption depends upon the nature of the adsorbent used at the same temperature. Moreover, the greater the surface area of the adsorbent, the greater is its adsorption capacity. Due to this reason, the substances like charcoal and silica gel are excellent adsorbent because they have high porous structures and hence have large surface areas. For the same reason, activated charcoal and finely divided solid substances are better adsorbents. (ii) Nature of the gas: Different gases are adsorbed to different extents by the same adsorbent at the same temperature. The easily liquefiable gases (like HCl, NH3, Cl2, etc.) are adsorbed more easily than the permanent gases (like H2, N2, O2, etc.). The ease of liquefaction of a gas depends upon its critical temperature (i.e., the minimum temperature above which a gas cannot be liquefied by applying even high pressure). The critical temperature of a gas is related to the intermolecular forces. Hence higher is the critical temperature, the more easily the gas is liquefied and hence more readily it is adsorbed. (iii) Temperature: As already discussed, adsorption is an exothermic process and it is accompanied by evolution of heat. Hence, according to Le Chatelier’s principle, it can be seen that increase of temperature decreases the rate of adsorption and vice versa. Condensation Gas (Adsorbate) + solid (Adsorbent) Gas Adsorbed on solid + Heat Evaporation

In case of physisorption, the amount of adsorption decreases with increase of temperature in accordance with Le Chatelier’s principle. But in case of chemisorption, it is observed that the amount of adsorption initially increases with increase in temperature and then decreases(Figure10.3). This is due to the fact that a chemisorption process requires some activation energy and hence the

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Surface Chemistry

x m

Physisorption

Temperature (a)

10.5

Chemisorption x m

Temperature (b)

Figure 10.3 Adsorption isobars for (a) Physisorption and (b) Chemisorption initial rise. Once the adsorption process commences, the amount of adsorption decreases due to its exothermic nature. (iv) Pressure: At constant temperature, the adsorption of a gas increases with increase of pressure according to Le Chatelier’s principle; since a dynamic equilibrium exists between the adsorbed gas and the adsorbent solid. (v) Activation of adsorbent: In order to increase adsorption, activation of the adsorbing power of the adsorbent is very necessary. This is done by increasing the surface area of the adsorbent, which can be achieved by the following ways: (a) By making the surface of the adsorbent rough through mechanical rubbing. (b) By subdividing the adsorbent into smaller pieces or grains. (c) By removing the gases already adsorbed by strong heating; due to this, its pores are opened, hence the rate of adsorption activity increases.

10.2.8 Adsorption Isotherms The extent of adsorption on a given surface increases with increase in pressure (for gases) and concentration (for solution) at constant temperature. The extent of adsorption is usually expressed as x/m, where m is the mass of the adsorbent and x is the mass of the adsorbate when adsorption equilibrium is reached. “A graph between the amount of the gas adsorbed per gram of the adsorbent (x/m) and the equilibrium pressure at constant temperature is called adsorption isotherm.” Freundlich Adsorption Isotherm To understand the effect of pressure on adsorption of a gas, we consider adsorption as an equilibrium process. When the adsorbent and adsorbate are enclosed in a closed vessel, after an initial decrease in the pressure of the gas, gas pressure as well as the amount of gas adsorbed reaches constant values. This is because when the equilibrium is attained, the rate of adsorption becomes equal to the rate of desorption (Figure 10.4).  x From the graph, it is clear that the extent of adsorption   increases with increasing pressure and  m becomes maximum at Ps, called the saturation pressure. At Ps, the rate of adsorption becomes equal to the rate of desorption (i.e., equilibrium is attained), and further increase in pressure does not alter the equilibrium.

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10.6 Engineering Chemistry

x = KP 0 m

x m

1 x = KP n m

x = KP 1 m

Ps (Saturation pressure)

Figure 10.4 Freundlich adsorption isotherm The following observations can be made from graph: (i) At low pressure, the graph is almost straight line and sloping. This is represented by the following equation: x ∝ P1 m or x = KP1 m (1)  x (ii) At high pressure,   is independent of pressure, that is, the graph becomes almost parallel to  m x-axis. This is represented by the following equation: x ∝ P0 m or x = K × P 0 (2) m  x (iii) In the intermediate range of pressure,   will depend on P raised to powers between 1 and 0,  m that is, fractions. So, for a small range of pressure values, we can write, 1

x ∝ Pn m or 1 x = KP n (3) m

Where n = positive integer and n and K are constants depending upon the nature of the adsorbate and adsorbent at a particular temperature. This relationship was originally put forward by Freundlich in 1909 and is known as Freundlich adsorption isotherm.

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Surface Chemistry

10.7

To test the validity of this equation, taking logarithms on both sides, we get and compare with straight line equation log

x 1 = log K + log P (4) m n

y = c + mx → Straight line

x against log P should, therefore, gives a straight line with slope equal to m 1 and ordinate intercept equal to log K (Figure 10.5). n A graph between log

log

1 = n ) m e(

Slo

x m

Intercept (c) = log K log P

Figure 10.5 Linear graph between log x and log P m Limitations of Freundlich adsorption isotherm: When the experiment’s values are plotted, it shows some deviation from linearity, especially at high pressure. Hence, this relation is suitable only at low pressure. Langmuir Adsorption Isotherm In 1916, Irving Langmuir proposed another adsorption isotherm, which explained the variation of adsorption with pressure. Based on his theory, he derived Langmuir equation, which depicted a relationship between the number of active sites of the surface undergoing adsorption and pressure. Assumption of Langmuir Isotherm (i) Fixed number of vacant or adsorption sites are available on the surface of a solid. (ii) All the vacant sites are of equal size and shape on the surface of the adsorbent. (iii) Each site can hold maximum of one gaseous molecule, and a constant amount of heat energy is released during this process. (iv) The ability of a molecule to adsorb at a given site is independent of the occupation of the neighboring sites. (v) Dynamic equilibrium exists between the adsorbed gaseous molecules and the free gaseous molecules. Adsorption A( g ) + B( s) AB Desorption

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10.8 Engineering Chemistry where A(g) = Unadsorbed gaseous molecule B(s) = Unoccupied metal surface AB = Adsorbed gaseous molecule (vi) Adsorption is monolayer or unilayer. Derivations of Langmuir Adsorption Equation (i) Calculation of equilibrium constant: Langmuir proposed the dynamic equilibrium that exists between adsorbed gaseous molecules and the free gaseous molecules. Using the equilibrium equation, equilibrium constant can be calculated: K

a AB A( g ) + B( s) K

Where Ka = equilibrium constant for forward reaction Kd = equilibrium constant for backward reaction According to kinetic theory, Rate of forward reaction = Ka[A][B] Rate of backward reaction = Kd[A][B] At equilibrium, Rate of forward reaction = Rate of backward reaction K a [ A][ B ] = K d [ AB] Ka [ AB ] K= = K d [ A][ B ] where K = adsorption constant or equilibrium constant (ii) Derivation of Langmuir equation: To derive Langmuir equation, a new parameter ‘θ’ is introduced. Let θ be the number of sites of the surface, which are covered with gaseous molecules. (1−θ) is the number of vacant sites. Initially, the rate of adsorption is high because the number of vacant sites is quite large compared to the filled sites. As adsorption process increases, the number of vacant sites for adsorption increases. We know that the rate of adsorption depends upon the pressure P and the number of vacant sites (1−θ). Rate of adsorption ∝ (1 − θ) ⋅ P (1) Rate of adsorption = K a (1 − θ) ⋅ P

Rate of desorption under the same conditions depends on θ, the faction of surface covered. Rate of desorption = K d θ (2)

At equilibrium, Rate of adsorption = Rate of desorption K a (1 − q ) P = K d ⋅q There

q =

Ka P Kd + Ka P

Divide numerator and denominator on RHS by Kd .

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Surface Chemistry

10.9

We get

Ka ⋅P Kd θ= Kd Ka + P Kd Kd KP = 1 + KP K where K = a Kd KP θ= 1 + KP This is known as Langmuir adsorption equation. (iii) Freundlich adsorption equation: A special case of Langmuir equation: We consider Langmuir Equation KP θ= 1 + KP (a) At low pressure, KP < < 1 Therefore θ = KP (1) or θ ∝ P1 (b) At high pressure, KP > > 1 KP θ= =1 KP (2) θ ∝ P0 Combining the result of equations (1) and (2), θ = KP 0 −1

or θ = KP n

(3)

Equation (3) is in agreement with Freundlich adsorption equation. Hence, we can say that Freundlich adsorption equation is a special case of Langmuir equation (Figure 10.6).

KP 1 + KP

θ=1 (Zero order)

θ θ = KP (Ist order) Pressure, P

Figure 10.6 Langmuir variation of coverage with pressure for molecular adsorption

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10.10 Engineering Chemistry (iv) Verification of Langmuir equation: The verification of Langmuir equation can be carried out as follows: The amount of gas adsorbed per unit mass of absorbent is x ∝ θ x = K1θ (4) By substituting the value of θ, K1 KP 1 + KP K2 P x= 1 + KP x=

where K 2 = K1K Dividing both sides by P and then taking reciprocals, it gives

P 1  K = P + x K 2  K 2  Since K and K 2 are constants for a given system. K Plotting a graph between P/x against P should give a straight line with a slope and interK 2 1 . Figure 10.7 shows a linear relationship for same experimental data. cept K2

P x

)= e (m

Slo

Intercept (c) =

K K2

1 K2

Figure 10.7 Linear graph of P/x versus P The Langmuir adsorption equation is only valid for unimolecular layer adsorption, not for multilayer adsorption in several solids. BET Adsorption Isotherm The concept of the theory is an extension of the Langmuir theory, which is a theory for monolayer molecular adsorption to multilayer adsorption. In 1938, Stephen Brunauer, Paul Hugh Emmett, and Edward Teller proposed this theory (BET theory) after their initials. BET Adsorption Isotherm is shown in Figure 10.8. The following hypothesis of BET theory: (i) Gas molecules adsorb on a solid in layer infinitely. (ii) There is no interaction between each adsorption layer. (iii) Langmuir theory can be applied to each layer. The resulting BET equation is expressed as follows:

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Surface Chemistry 10.11

V/Vm

Figure 10.8 BET adsorption isotherm

C −1  P  1 1 = +    P   VmC  P0  VmC V  0  − 1  P  

(1)

1 1 (C − 1) ⋅x+ = (2) VmC VmC  1   V  − 1   x  

Here P and P0 = equilibrium and saturation pressure of adsorbates at the temperature of adsorption V = Volume of gas adsorbed Vm = Volume of monolayer absorbed quantity C = BET constant  E − EL  C = exp  1 (3)  RT 

Where E1 = Heat of adsorption for the first layer EL = Heat of liquefaction for the second and higher layers Verification of BET Equation Equation (1) is an adsorption isotherm and can be plotted as a straight line (Figure 10.9). Plotting of P against P/P0 should be a straight line. V ( P0 − P ) C −1 1 and intercept with the value of . Thus, by knowing the The slope of the line is given as VmC Vm ⋅ C value of slope and intercept, both Vm and C can be evaluated by using the following equation: Vm =

1 (4) A+ I

and

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10.12 Engineering Chemistry

C = 1+

A I

P V(P0−P )

(5)

C−1 Slope = V m⋅C (A)

1 Intercept = V C m (I ) P/P0

Figure 10.9 Linear graph of P/P0 Vs P/V(P0-P) Derivation of BET Equation Consider a surface and molecules that are adsorbed on different layers, that is, adsorption on multilayer (Figure 10.10). Consider θ0, θ1, θ2 ,……… θn = surface area covered by 0,1, 2,……, n layers of adsorbed molecules. K

1 A( g ) + S AS K −1

K2 A( g ) + AS A2 S K

−2

K3 A( g ) + A2 S A3 S and so on. K −3

4th layer 2

3rd layer

layer 1st layer

Surface

Figure 10.10 Adsorption of gas molecules on the surface of a solid At equilibrium, θ0 must remain constant.

∴ Rate of evaporation from first layer = Rate of condensation onto base surface K −1θ1 = K1 Pθ0 Similarly, at equilibrium, θ1 must remain constant.

(1)

 Rate of condensation on the base surface  Rate of condensation on first layer  =  ∴ + +  Rate of evaporation from the seond layer   Rate of evaporation from first layer      K1 Pθ0 + K −2 θ2 = K 2 Pθ1 + K −1θ1

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Surface Chemistry 10.13

Substituting in equation (1) gives

And so on. For other layers,

K −2 θ2 = K 2 Pθ1 K − i θi = K i Pθi −1

(2)

According to the definitions, total surface area of the catalyst, ∞

A = ∑ θi (3)

i=0

Total volume of the gas adsorbed on the surface, ∞

V = V0 ∑ iθi

(4)

i=0

Where V0 = Volume of the gas adsorbed on one square centimeter of surface when it is covered with a complete layer. Then, ∞

V V = = AV0 Vm

∑ iθi i =0 ∞

∑ θi

(5)

i =0

Vm = Volume of the gas adsorbed when the entire surface is covered with a complete monolayer From equation (1),  K  θ1 =  1  Pθ0  K −1  = Y θ0

 K   Where Y =  1  P   K −1   By assuming that the properties of 1st, 2nd, and nth layers are equivalent, K −2 K −3 K − i = … = g (6) K2 K3 Ki

Similarly,

 K  θ2 =  2  Pθ1  K −2  = X θ1

 P Here  X =  g 

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10.14 Engineering Chemistry Generally, θi = X θi −1 = X i −1θ1 = X i −1Y θ0 = CX i θ0

{} Y X

Where C = Substituting in equation (5), ∞

V = Vm

∑ iθi i=0 ∞

∑ θi

∞

C θ0 ∑ iX i i=0

(θ0 + θ1 + θ2 + ........)

i=0

∞

C θ0 ∑ iX i i =1

∞   θ0 1 + C ∑ X i    ∞

i =1

C ∑ iX i i =1

∞   + C Xi  1 ∑   i =1

(7)

On solving, V CX = Vm (1 − X )(1 − X + CX )

At saturation, pressure of gas P0, an infinite number of adsorbate layers must build on the surface. This means at P0, X must equal 1. ∴ g = P0

∴ X =

P P0

Substituting in equation (7), we get BET isotherm equation VmCX (1 − X ){1 + (C − 1)X } X 1 X (C − 1) = + V (1 − X ) VmC VmC P 1 C −1 P × = + V (P0 − P ) VmC VmC P0 V=

Where P = Pressure at the absorption equilibrium P0 = Saturated vapour pressure of the adsorbent at different temperatures

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Surface Chemistry 10.15

Calculation of Surface Area of Adsorbent by BET Method BET method is widely used for the calculation of surface areas of solids by physical adsorption of gas molecules. The total surface area STotal and a specific surface area S are evaluated by the following equations: Vm NS V

SBET =

STotal = SBET × a V NS = m a V

(1)

Where N = Avogadro’s number S = Adsorption cross section V = Molar volume of adsorbent gas Vm = Volume of monolayer adsorbed gas a = Molar weight of adsorbed species For example, to determine the inner surface of a hardened cement paste, by plotting a graph between the quantities of adsorbed water at different levels of relative humidity, a BET plot is obtained. From the slope A and intercept I, one can calculate Vm and C by using the following equation: 1 A+ I A C = 1 + , where A = 24.20, I = 0.33 I 1 1 Vm = = A + I 24.20 + 0.33 = 0.0408 g/g Vm =

SBET can be calculated by using the equation, S BET =

Vm NS V

The surface covered by one water molecule, S = 0.114 nm 2 So, 1 × 6.023 × 10 23 18 = 155.63 × 10 −5 × 10 23 nm 2 = 155.63 × 1018 nm 2 /g

S BET = 0.0408 × 0.114 ×

= 155.63 m 2 /g Surface area of the hardened cement = 155.63 m 2/g It means the cement paste has inner surface of 155.63 m 2 per g of cement.

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10.16 Engineering Chemistry

10.2.9 Applications of Adsorption Adsorption finds extensive applications both in industry and research laboratory. A few applications are briefly described below: (i) In preserving vacuum: Adsorption process is used in the production of vacuum by using cultivated charcoal in Dewar’s flask. (ii) In clarification of sugar: Sugar is decolourized by treating sugar solution with charcoal powder. (iii) In gas masks: All gas masks are devices containing suitable adsorbent so that the poisonous gases present in the atmosphere are preferentially adsorbed and air for breathing is purified. (iv) In heterogeneous catalysis: Adsorption is useful in heterogeneous catalysis. This theory is called adsorption theory. According to this theory, gaseous reactants are adsorbed on the surface of the solid. As a result, concentration of reactant increases. Hence, the rate of reaction increases. For example, in the manufacturing of ammonia using iron as catalysts, V2O5 is used as catalyst in the manufacturing of H2SO4 by contact process. (v) In softening of hard water: In water-softening process, hard water is passed through zeolite bed in which Ca2+ and Mg2+ ions of hard water are replaced by equal amount of Na+ ions. So by this process, water becomes soft. (vi) In adsorption indicator: Various dyes, which have the property of adsorption, have been intruded particularly in precipitation titration. For example, by the use of eosin as indicator, KBr is easily titrated with AgNO3. (vii) In dehydration process: Silica gel and alumina gel adsorbents are used to remove moisture and dehumidifies the air from rooms and instruments. (viii) In chromatographic analysis: The selective adsorption of certain substances from a solution by a particular solid adsorbent filled in a long and wide vertical tube: column chromatography. Examples are as follows: (a) Separation of isomeric aromatic or aliphatic hydrocarbons (b) Separation of two or more components of mixture (c) Purification of substances from their contaminants (d) Identification of biochemicals such as amino acids, neutral lipids, carbohydrates, etc. (ix) In Froth floatation process: In this process, finely divided sulphide ores (PbS, ZnS, Cu2S, etc.) are mixed with pine oil and agitated with water containing detergent. Air is then bubbled through this mixture, and air bubbles stabilized by the detergent adsorb mineral particles, which float to the surface while the silica impurities settle down. (x) In medicine: For the treatment of arsenic poisoning, colloidal Fe(OH)3 is administered as an antidote in which Fe(OH)3 adsorbs arsenic and retains it and thus can be removed from the body by vomiting. (xi) In dyeing: During the process of mordant dyeing, the metal ions like Al3+, Fe3+, and Cr6+ adsorb the dye, which otherwise does not adhere to the fabric. (xii) Ion exchange adsorption process is used to remove excess of sodium salt from the body fluids of a patient by providing him/her to a suitable cation exchanger to consume. Similarly, anionic resin is used to neutralize the excess acid in the stomach. (xiii) By adsorption, we can easily remove polluting gases and control indoor pollution. (xiv) By the use of surface active agents in detergents, paints, lubrication, surface cleaner, etc., surface active agents displace the adhering impurities from all surfaces by the phenomenon of wetting.

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Surface Chemistry 10.17

(xv) Soil contains small amounts of colloidal fractions in the form of very fine particles of clay, which adsorb moisture in which essential plant nutrients N, P, and K dissolve and are taken up by the plants by capillary action.

10.3 COLLOIDAL STATE The origin of the study of this field lies with Thomas Graham, who in 1861 studied the diffusion of liquids and solutions across the animal membranes, and as a result, he divided the soluble substances into two classes: (i) Crystalloids diffuse readily in solution and those solution that can readily pass through animal or vegetable membranes, for example, urea, sugar, salts, acids, bases, and other crystalline substances. (ii) Colloids (from the Greek words “kola” and “eidos”, which mean glue and like, respectively), diffuse very slowly in solution and those solution that cannot pass through animal or vegetable membranes, for example, glue, starch, albumin, proteins, and other amorphous substances. In recent years, this classification of substances has undergone a great change because distinction between crystalloids and colloids was not rigid. In fact, any substance, regardless of its nature, could be converted into a colloid by subdividing it into particles of small size (1–1000 nm).

10.3.1 Types of Solution On the basis of the size of the particle, the solutions may be of the following three types: (i) True solution: True solution is a homogeneous solution containing dispersed particles of molecular size (less than 10−9 m or 1 nm), for example, NaCl or glucose solution in water. (ii) Suspension solution: Suspension is a heterogeneous solution containing mixture of suspended insoluble particles of size (greater than 10000 A° or 1000 nm). (iii) Colloidal solution: Colloidal solution is a heterogeneous solution in which a substance is distributed in colloidal state (1 nm–1000 nm) in an insoluble medium. The common characteristics among the three solutions are given in Table 10.3. Table 10.3 Characteristics of true solution, colloidal solution, and suspension S. no.

Property

True solutions

Colloidal solutions

Suspension solutions

1. 2.

State Particle size

Homogeneous Less than 1 nm

Heterogeneous More than 1000 nm

Invisible

4. 5.

Visibility of particle Diffusion Filterability

6. 7.

Settling Appearance

Heterogeneous Between 1 nm and 1000 nm Visible under ultramicroscope Diffuse slowly Pass through only ordinary filter paper, not through animal membrane. Do not settle Translucent

Diffuse quickly Pass through ordinary filter paper as well as animal membrane. Do not settle Clear and transparent

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Visible Do not diffuse Do not pass through ordinary filter paper as well as animal membrane. Settle on standing Opaque

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10.18 Engineering Chemistry

10.3.2 Classification of Colloids In colloidal system, the terms dispersed phase and dispersion medium are used. Dispersed phase means the substance distributed in the dispersion medium in the form of colloidal particles, and dispersion medium means the medium in which the substance is dispersed in the form of colloidal particles. Just as in solution, where solute and solvent are used, in colloidal system, dispersed phase and dispersion medium are used. Colloidal systems are classified in different ways: (i) Based on the physical state of the dispersed phase and dispersion medium: Depending upon the physical state of the dispersed phase and dispersion medium, that is, solids, liquids, or gases, eight types of colloidal systems are possible (Table 10.4). Table 10.4 Physical states of the dispersed phase and dispersed medium with examples S. no.

Dispersed phase

Dispersion medium

Colloidal system

Solid

Solid Sol

Solid

Liquid

Sol

3. 4.

Solid Liquid

Gas Solid

Solid aerosol Gel

5. 6. 7.

Liquid Liquid Gas

Liquid Gas Solid

Emulsion Liquid Aerosol Solid foam

Gas

Liquid

Foam or froth

Examples Some coloured glasses, gems, stones, minerals, etc. Some paints, cell fluids, muddy water, white of an egg, etc. Smoke, dust, fume, etc. Cheese, butter, jellies, curd, boot polish, etc. Milk, cream, medicine, etc. Fog, mist, cloud, spray, etc. Biscuit, cake, bread dough, floating soaps, etc. Foam, whipped cream, soap lather, detergent suds, etc.

(ii) Based on the nature of the dispersion medium: Depending upon the nature of the dispersion medium, the sols (solid in liquid) are given special names as in the following Table 10.5: Table 10.5 Nature of the dispersed medium and the names of the sols Dispersion medium

Name of the sol

Water Alcohol Benzene Gases

Aquasol or Hydrosol Alcosol Benzosol Aerosol

(iii) Based on the nature of interaction between dispersed phase and dispersion medium: On this basis, colloidal sols are divided into two categories, namely, lyophilic and lyophobic. If water is the dispersion medium, the terms used are hydrophilic and hydrophobic. (a) Lyophilic sols are those in which the dispersion medium possesses great affinity (liquidloving) for the dispersed phase. Examples are starch, glue, gelatin, agar sols in water, etc. Characteristics: (1) Lyophilic sols are directly formed colloidal sols; that is why they are also called intrinsic colloids.

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Surface Chemistry 10.19

(2) The dispersion medium is easily separated from the dispersed phase (by evaporation), and the sol can be made by simply remixing with dispersion medium and shaking. That is why, these sols are also called reversible sols. (3) These sols are quite stable and cannot be easily precipitated. (b) Lyophobic sols are those in which there is no affinity or interaction between the dispersion medium and the dispersed phase. Examples are gold sol, silver sol, and metal sulphide sol in water. Characteristics: (1) Their colloidal sols are generally prepared by indirect methods; that’s why they are called extrinsic colloids. (2) Such types of colloidal sols are not precipitated out once they formed. These sols are also called irreversible sols. (3) These sols are easily precipitated (or coagulated), and hence are not stable. The points of difference between lyophilic sols and lyophobic sols are given below in Table 10.6: Table 10.6 Differences between lyophilic sols and lyophobic sols S. no.

Property

Lyophilic sols

Ease of preparation

Easily prepared by directly mixing with the liquid dispersion medium. They are quite stable and are not easily precipitated or coagulated.

2. 3. 4.

5. 6. 7. 8. 9. 10.

Lyophobic sols

Cannot be prepared directly. It can be prepared by special methods only. Stability They are easily precipitated by addition of small amount of electrolyte. Hydration They are highly hydrated. They are not much hydrated. Nature of the particles Particles are solvent-loving and Particles are solvent-hating in the form of single molecule. and consist of large number of associated molecules. Viscosity Viscosity is higher than that of Viscosity is almost the same as the solvent. that of the solvent. Visibility The particles cannot be The particles can be detected readily, even under readily detected under ultramicroscope. ultramicroscope. Stability Very stable Poor stability of particle Surface tension Their surface tension is lower Their surface tension is nearly than that of the dispersion same as that of the dispersion medium. medium. Reversibility Reversible in nature Irreversible in nature Action of electrolyte Coagulation can be brought about The addition of even small only by the addition of large quantity of electrolyte can cause quantities of electrolytes. coagulation.

(iv) Based on the type of particles of the dispersed phase: Depending upon the different size of the colloids, colloidal dispersion may be divided into the following three categories: (a) Multimolecular colloids: Multimolecular colloids are colloidal solutions in which colloidal particles (or dispersed phase) consist of aggregates of atoms or molecules, each having diameter less than 1 nm. In these aggregates, the atoms or molecules are held together by weak force of attraction. They are lyophobic in nature. For example,

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10.20 Engineering Chemistry

(1) A gold sol in water consists of dispersed particles of various sizes made up of several atoms of gold. (2) A sulphur sol in water consists of dispersed particles containing a large number of S8 molecules. (b) Macromolecular colliods: Macromolecular colloids are colloidal solutions in which the dispersed particles are very large molecules of high molecular mass (called macromolecules). These particles contain a large number of atoms (joined together through covalent bonds), each having dimensions compared to those of colloidal state. They are lyophilic sols and possess gold number. For example, (1) Naturally occurring macromolecules such as starch, cellulose, proteins, enzymes and gelatine, etc. (2) Man-made macromolecules like polyethylene, nylon, polystyrene, synthetic rubber, etc. (c) Associated colloids–micelles: Associated colloids are the colloidal solutions in which the colloidal particles behave as normal, strong electrolytes when dissolved in a medium at low concentrations, but at higher concentrations, they form aggregated particles of colloidal dimensions, called micelles. The formation of micelles take

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